diff --git a/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/3568.Minimum-Moves-to-Clean-the-Classroom.cpp b/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/3568.Minimum-Moves-to-Clean-the-Classroom.cpp
new file mode 100644
index 000000000..7a5307a84
--- /dev/null
+++ b/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/3568.Minimum-Moves-to-Clean-the-Classroom.cpp
@@ -0,0 +1,74 @@
+using state = tuple<int,int,int,int>;
+class Solution {
+public:
+ int minMoves(vector<string>& grid, int energy) {
+ int m = grid.size(), n = grid[0].size();
+
+ pair<int,int>start;
+ vector<pair<int,int>>litter_pos;
+ vector<vector<int>>litter_idx(m, vector<int>(n,-1));
+ 
+ for (int i=0; i<m; i++) + for (int j=0; j<n; j++) { + if (grid[i][j]=='S') + start = {i,j}; + else if (grid[i][j]=='L') + { + litter_pos.push_back({i,j}); + litter_idx[i][j] = litter_pos.size()-1; + } + } + int L = litter_pos.size(); + + vector<vector<vector<vector<bool>>>> visited(
+ m, vector<vector<vector<bool>>>(
+ n, vector<vector<bool>>(energy + 1, vector<bool>(1 << L, false)))); + + visited[start.first][start.second][energy][0] = true; + + vector<pair<int,int>>dir={{0,1},{0,-1},{1,0},{-1,0}};
+
+ queue<state> q;
+ q.push({start.first, start.second, energy, 0});
+ int step = 0;
+
+ while (!q.empty())
+ {
+ int len = q.size();
+ while (len--)
+ {
+ auto [x,y,e,mask] = q.front();
+ q.pop();
+ if (mask==(1<<l)-1) return step; + + for (int k=0; k<4; k++) + { + int nx = x+dir[k].first; + int ny = y+dir[k].second; + if (nx<0&#124;&#124;nx>=m&#124;&#124;ny<0&#124;&#124;ny>=n) continue;
+ 
+ char cell = grid[nx][ny];
+ if (cell=='X') continue;
+ 
+ int newEnergy = e-1;
+ if (newEnergy < 0) continue; + if (cell == 'R') newEnergy = energy; + + int newMask = mask; + if (grid[nx][ny]=='L') + { + int idx = litter_idx[nx][ny]; + newMask &#124;= (1<<idx); + } + if (!visited[nx][ny][newEnergy][newMask]) { + visited[nx][ny][newEnergy][newMask] = true; + q.emplace(nx, ny, newEnergy, newMask); + } + } + } + step++; + } + + return -1; + } +}; diff --git a/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/Readme.md b/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/Readme.md new file mode 100644 index 000000000..a8187eb19 --- /dev/null +++ b/BFS/3568.Minimum-Moves-to-Clean-the-Classroom/Readme.md @@ -0,0 +1,5 @@ +### 3568.Minimum-Moves-to-Clean-the-Classroom + +典型的BFS,但是我们需要定义一个composite state来避免重复。本题中因为最有路径可能需要重复经过相同的cell,所以我们还需要记录每一步时当前的energy,已经访问过的litter。这样所有的状态种类有`20*20*50*2^10=2e7`,复杂度勉强够用。 + +具体的做法就是常规的BFS。每次从队列中弹出[x,y,energy,mask],其中mask是一个二进制数,用bit位来记录哪些编号的垃圾已经被清理。从当前[x,y]往四个方向尝试移动,每次移动需要更新energy(减一,或者遇到R就reset),也可能需要更新mask(即遇到L)。在将新状态加入队列之前,查看一下这个state之前是否已经加入过队列(即可以在更早的时间被访问到)以避免重复搜索。 diff --git a/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II.cpp b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II.cpp new file mode 100644 index 000000000..440611fad --- /dev/null +++ b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II.cpp @@ -0,0 +1,80 @@ +using ll = long long; +const int MAXN = 100000; +const int LOGN = 17; +class Solution { +public: + vector<pair<int,int>> adj[MAXN];
+ int up[MAXN][LOGN+1];
+ int depth[MAXN];
+ ll distRoot[MAXN];
+
+ void dfs(int cur, int parent)
+ {
+ up[cur][0] = parent;
+ for(auto &[v,w]: adj[cur])
+ {
+ if(v == parent) continue;
+ depth[v] = depth[cur] + 1;
+ distRoot[v] = distRoot[cur] + w;
+ dfs(v, cur);
+ }
+ }
+
+ int lca(int a, int b)
+ {
+ if(depth[a] < depth[b]) swap(a,b); + int diff = depth[a] - depth[b]; + for(int k = 0; k <= LOGN; k++){ + if(diff & (1<<k)) a = up[a][k]; + } + if(a == b) return a; + for(int k = LOGN; k>= 0; k--){
+ if(up[a][k] != up[b][k]){
+ a = up[a][k];
+ b = up[b][k];
+ }
+ }
+ return up[a][0];
+ }
+
+ ll dist(int a, int b)
+ {
+ int c = lca(a,b);
+ return distRoot[a] + distRoot[b] - 2*distRoot[c];
+ }
+ 
+ vector<int> minimumWeight(vector<vector<int>>& edges, vector<vector<int>>& queries) 
+ {
+ int n = edges.size()+1;
+ 
+ for (int i = 0; i < n-1; i++) + { + int u = edges[i][0], v = edges[i][1], w = edges[i][2]; + adj[u].push_back({v,w}); + adj[v].push_back({u,w}); + } + + depth[0] = 0; + distRoot[0] = 0; + dfs(0, 0); + + for(int k = 1; k <= LOGN; k++) { + for(int v = 0; v < n; v++) { + up[v][k] = up[up[v][k-1]][k-1]; + } + } + + vector<int>rets;
+ for (auto& q: queries)
+ { 
+ int u = q[0], v = q[1], w = q[2]; 
+ ll d_uv = dist(u,v);
+ ll d_vw = dist(v,w);
+ ll d_uw = dist(u,w);
+ ll ans = (d_uv + d_vw + d_uw) / 2;
+ rets.push_back(ans);
+ }
+ 
+ return rets;
+ }
+};
diff --git a/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md
new file mode 100644
index 000000000..21b7d3d06
--- /dev/null
+++ b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md
@@ -0,0 +1,3 @@
+### 3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II
+
+本题的第一个知识点是:在一棵树里,联通u,v,w三个节点的最小子树的权重和,就是`[dist(u,v)+dist(u,w)+dist(v,w)]/2`.
diff --git a/Readme.md b/Readme.md
index 2cc7165b0..f62e672a9 100644
--- a/Readme.md
+++ b/Readme.md
@@ -107,6 +107,7 @@
 [2836.Maximize-Value-of-Function-in-a-Ball-Passing-Game](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2836.Maximize-Value-of-Function-in-a-Ball-Passing-Game) (H) 
 [2846.Minimum-Edge-Weight-Equilibrium-Queries-in-a-Tree](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2846.Minimum-Edge-Weight-Equilibrium-Queries-in-a-Tree) (H) 
 [2851.String-Transformation](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2851.String-Transformation) (H+) 
+[3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II) (H) 
 * ``Binary Search by Value`` 
 [410.Split-Array-Largest-Sum](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/410.Split-Array-Largest-Sum) (H-) 
 [774.Minimize-Max-Distance-to-Gas-Station](https://github.com/wisdompeak/LeetCode/tree/master/Priority_Queue/774.Minimize-Max-Distance-to-Gas-Station) (H) 
@@ -621,7 +622,8 @@
 [913.Cat-and-Mouse](https://github.com/wisdompeak/LeetCode/tree/master/BFS/913.Cat-and-Mouse) (H+) 
 [1728.Cat-and-Mouse-II](https://github.com/wisdompeak/LeetCode/tree/master/BFS/1728.Cat-and-Mouse-II) (H+) 
 [1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination](https://github.com/wisdompeak/LeetCode/tree/master/BFS/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination) (H-) 
-[1928.Minimum-Cost-to-Reach-Destination-in-Time](https://github.com/wisdompeak/LeetCode/tree/master/BFS/1928.Minimum-Cost-to-Reach-Destination-in-Time) (H-) 
+[1928.Minimum-Cost-to-Reach-Destination-in-Time](https://github.com/wisdompeak/LeetCode/tree/master/BFS/1928.Minimum-Cost-to-Reach-Destination-in-Time) (H-) 
+[3568.Minimum-Moves-to-Clean-the-Classroom](https://github.com/wisdompeak/LeetCode/tree/master/BFS/3568.Minimum-Moves-to-Clean-the-Classroom) (H-) 
 * ``拓扑排序`` 
 [207.Course-Schedule](https://github.com/wisdompeak/LeetCode/tree/master/BFS/207.Course-Schedule) (H-) 
 [210.Course-Schedule-II](https://github.com/wisdompeak/LeetCode/tree/master/BFS/210.Course-Schedule-II) (M) 
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