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Commit 77f6dc2

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‎Advanced Graphs/Connectedhorses.cpp‎

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/*
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Name: Mehul Chaturvedi
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IIT-Guwahati
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*/
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/*
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You all must be familiar with the chess-board having
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8*8 squares of alternate black and white cells. Well, here
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we have for you a similar
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N*M size board with similar arrangement of black and white
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cells.
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A few of these cells have Horses placed over them.
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Each horse is unique. Now these horses are not the
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usual horses which could jump to any of the
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8 positions they usually jump in. They can move only if there is another horse on one of the 8-positions that it can go to usually and then both the horses will swap their positions. This swapping can happen infinitely times.
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A photographer was assigned to take a picture of all the different ways that the horses occupy the board! Given the state of the board, calculate answer. Sincethis answer may be quite large, calculate in modulo
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10^9+7
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Input:
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First line contains
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T which is the number of test cases.
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T test cases follow first line of each containing three integers
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N, M and Q where
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N,M is the size of the board and
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Q is the number of horses on it.
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Q lines follow each containing the 2 integers
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X and Y which are the coordinates of the Horses.
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Output:
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For each test case, output the number of photographs taken by photographer.
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Constraints:
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1<=T<=10
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1<=N,M<=1000
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1<=Q<=N*M
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SAMPLE INPUT
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2
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4 4 4
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1 1
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1 2
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3 1
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3 2
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4 4 4
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1 1
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1 2
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3 1
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4 4
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SAMPLE OUTPUT
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4
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2
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*/
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#include <bits/stdc++.h>
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using namespace std;
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#define pii pair<ll,ll>
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#define MOD 1000000007
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#define ll long long
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ll dfs(ll i, ll j, vector<pii>** graph, ll** visited){
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visited[i][j] = 1;
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ll answer = 1;
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for(ll k = 0; k < graph[i][j].size(); k++){
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ll x = graph[i][j][k].first;
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ll y = graph[i][j][k].second;
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if(!visited[x][y]){
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answer = (answer+dfs(x, y, graph, visited))%MOD;
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}
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}
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return answer;
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// ll count = 0;
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// for (ll i = 0; i < graph[sx][sy].size(); ++i)
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// {
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// ll left = (graph[sx][sy].at(i)).first;
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// ll right = (graph[sx][sy].at(i)).second;
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// if (visited[left][right] == 0)
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// {
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// visited[left][right] = 1;
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// count = (count+1)%MOD;
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// }
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// }
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// return count%MOD;
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}
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void fill_vector(vector<pii>** graph, ll** board, ll m, ll n){
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for (ll i = 0; i < n; ++i)
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{
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for (ll j = 0; j < m; ++j)
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{
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if (board[i][j] == 1)
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{
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graph[i][j].push_back(make_pair(i, j));
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if (i+2<n && j+1<m && board[i+2][j+1] == 1)
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{
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//return ;
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graph[i][j].push_back(make_pair(i+2, j+1));
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}
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//return ;
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if (i+2<n && j-1>=0 && board[i+2][j-1] == 1)
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{
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graph[i][j].push_back(make_pair(i+2, j-1));
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}
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if (i-2>=0 && j+1<m && board[i-2][j+1] == 1)
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{
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graph[i][j].push_back(make_pair(i-2, j+1));
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}
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if (i-2>=0 && j-1>=0 && board[i-2][j-1] == 1)
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{
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graph[i][j].push_back(make_pair(i-2, j-1));
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}
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if (i+1<n && j+2<m && board[i+1][j+2] == 1)
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{
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graph[i][j].push_back(make_pair(i+1, j+2));
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}
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if (i+1<n && j-2>=0 && board[i+1][j-2] == 1)
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{
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graph[i][j].push_back(make_pair(i+1, j-2));
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}
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if (i-1>=0 && j+2<m && board[i-1][j+2] == 1)
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{
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graph[i][j].push_back(make_pair(i-1, j+2));
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}
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if (i-1>=0 && j-2>=0 && board[i-1][j-2] == 1)
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{
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graph[i][j].push_back(make_pair(i-1, j-2));
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}
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}
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}
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}
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return;
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}
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int main( int argc , char ** argv )
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{
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ios_base::sync_with_stdio(false) ;
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cin.tie(NULL) ;
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//To get the factorial
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ll *factorial = new ll[1000000];
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factorial[1] = 1;
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for(ll i = 2; i < 1000000; i++){
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factorial[i] = (factorial[i-1]*i)%MOD;
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}
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//Main part of the input
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ll t;
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cin>>t;
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while(t--){
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ll n, m;
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ll q;
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cin>>n>>m>>q;
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//Create a 2d array for board
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ll** board = new ll*[n];
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for (ll i = 0; i < n; ++i)
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{
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board[i] = new ll[m];
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for (ll j = 0; j < m; ++j)
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{
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board[i][j] = 0;
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}
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}
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//Fill the board
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while(q--){
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ll x, y;
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cin>>x>>y;
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board[x-1][y-1] = 1;
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}
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//Creating a graph 2d array
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vector<pii>** graph = new vector<pii>*[(n)];
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for (ll i = 0; i < n; ++i)
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{
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graph[i] = new vector<pii>[m];
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}
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//Fill the graph
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fill_vector(graph, board, m, n);
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// for (ll i = 0; i < n; ++i)
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// {
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// for (ll j = 0; j < m; ++j)
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// {
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// if (graph[i][j].size()!=0)
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// {
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// for (ll k = 0; k < graph[i][j].size(); ++k)
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// {
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// cout<<graph[i][j][k].first<<" "<<graph[i][j][k].second<<" ";
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// }
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// cout <<'\n';
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// }
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// }
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// }
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//Count and multiply the number of elements
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ll** visited = new ll*[n];
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for (ll i = 0; i < n; ++i)
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{
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visited[i] = new ll[m];;
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for (ll j = 0; j < m; ++j)
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{
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visited[i][j] = 0;
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}
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}
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ll ans = 1;
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for (ll i = 0; i < n; ++i)
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{
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for (ll j = 0; j < m; ++j)
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{
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if (visited[i][j] == 0 && board[i][j] == 1)
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{
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// cout << i<<" "<<j<<" "<<" "<<dfs(i, j, graph, visited, m, n) << '\n';
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// cout << "Here "<<factorial[2] << '\n';
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// cout << i<<" "<<j<<" "<<" "<<dfs(i, j, graph, visited, m, n) << '\n';
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ans = (ans*factorial[dfs(i, j, graph, visited)])%MOD;
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}
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}
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}
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cout << ans << '\n';
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}
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return 0 ;
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}

‎Advanced Graphs/Permutationsswap.cpp‎

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/*
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Name: Mehul Chaturvedi
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IIT-Guwahati
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*/
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/*
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Kevin has a permutation P of N integers 1, 2, ..., N, but he doesn't like it. Kevin wants to get a permutation Q.
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Also he believes that there are M good pairs of integers (ai , bi). Kevin can perform following operation with his permutation:
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Swap Px and Py only if (x, y) is a good pair.
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Help him and tell if Kevin can obtain permutation Q using such operations.
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Input format:
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The first line of input will contain an integer T, denoting the number of test cases.
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Each test case starts with two space-separated integers N and M. The next line contains N space-separated integers Pi. The next line contains N space-separated integers Qi. Each of the next M lines contains two space-separated integers ai and bi.
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Output format:
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For every test case output "YES" (without quotes) if Kevin can obtain permutation Q and "NO" otherwise.
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Constraints:
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1 ≤ T ≤ 10
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2 ≤ N ≤ 100000
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1 ≤ M ≤ 100000
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1 ≤ Pi, Qi ≤ N. Pi and Qi are all distinct.
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1 ≤ ai < bi ≤ N
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SAMPLE INPUT
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2
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4 1
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1 3 2 4
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1 4 2 3
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3 4
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4 1
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1 3 2 4
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1 4 2 3
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2 4
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SAMPLE OUTPUT
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NO
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YES
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*/
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#include <bits/stdc++.h>
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using namespace std;
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void dfs(vector<int> *graph, int start, bool *isVisited, unordered_set<int> *part){
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isVisited[start] = true;
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part->insert(start);
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for(int i = 0; i < graph[start].size(); i++){
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int next = graph[start][i];
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if(!isVisited[next]){
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dfs(graph, next, isVisited, part);
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}
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}
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}
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unordered_set< unordered_set<int> *> *getComponent(vector<int> *graph, int n){
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bool *isVisited = new bool[n];
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for(int i = 0; i < n; i++) isVisited[i] = false;
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unordered_set< unordered_set<int> * > *component = new unordered_set< unordered_set<int> * >();
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for(int i = 0; i < n; i++){
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if(!isVisited[i]){
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unordered_set<int> *part = new unordered_set<int>();
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dfs(graph, i, isVisited, part);
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component->insert(part);
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}
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}
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return component;
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}
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int main(){
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int flag;
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int t;
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cin >> t;
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while(t--){
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flag = 1;
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int n, m;
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cin >> n >> m;
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int *p = new int[n];
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for(int i = 0; i < n; i++){
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int x;
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cin >> x;
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p[i] = x-1;
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}
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int *q = new int[n+1];
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for(int i = 0; i < n; i++){
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int x;
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cin >> x;
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q[i] = x-1;
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}
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vector<int> *graph = new vector<int>[n];
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for(int i = 0; i < m; i++){
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int a, b;
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cin >> a >> b;
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graph[a-1].push_back(b-1);
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graph[b-1].push_back(a-1);
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}
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unordered_set< unordered_set<int> * > *component = getComponent(graph, n);
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// for(unordered_set< unordered_set<int> * >::iterator i = component->begin(); i != component->end(); i++){
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// for(unordered_set<int>::iterator j = (*i)->begin(); j != (*i)->end(); j++){
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// cout << (*j)+1 << " ";
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// }
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// cout << endl;
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// }
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for(unordered_set< unordered_set<int> * >::iterator i = component->begin(); i != component->end(); i++){
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set<int> setA;
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set<int> setB;
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for(unordered_set<int>::iterator j = (*i)->begin(); j != (*i)->end(); j++){
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setA.insert(p[*j]);
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setB.insert(q[*j]);
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}
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if(setA != setB){
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flag = 0;
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cout << "NO" << endl;
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break;
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}
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}
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if(flag) cout << "YES" << endl;
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}
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}

‎Advanced Graphs/Test.txt‎

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1
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2 7 12
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1 2
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2 2
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2 6
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1 7
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1 1
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2 3
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1 3
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1 4
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2 1
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2 5
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2 4
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1 6

‎Advanced Graphs/a.out‎

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