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Commit 3ae3ace

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1 parent d34314f commit 3ae3ace

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#include<bits/stdc++.h>
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using namespace std;
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typedef unsigned short ushort;
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const int maxSize = 25,inf=10000;
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ushort rect[maxSize][maxSize];
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//ushort rect2[maxSize][maxSize][maxSize][maxSize];
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auto rect2=new ushort[maxSize][maxSize][maxSize][maxSize];
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class Solution {
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public:
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int tilingRectangle(int n, int m) {
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if (n > m)swap(n, m);
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if (rect[n][m])return rect[n][m];
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//方形情况
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if (n == m)return rect[n][m] = 1;
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int ans=inf;
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//分解为方形+矩形(方形顶满情况)
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//int ans = 1 + tilingRectangle(n, m - n);
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//分解为方形+L形,枚举方形的边长,其中加的1就是方形占用数量
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/*for (int i = n-1; i > 0; i--)
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{
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//L形的分解可以在横轴也可以在竖轴枚举,下面的Rect2只枚举横轴,竖轴通过旋转改为枚举横轴
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//L形枚举
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ans = min(ans, solveL(n - i, m, n, m - i) + 1);
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}*/
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// decompose into R+R
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for (int i=1;i<n;++i)ans=min(ans,tilingRectangle(i,m)+tilingRectangle(n-i,m));
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for (int i=1;i<m;++i)ans=min(ans,tilingRectangle(n,i)+tilingRectangle(n,m-i));
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// decompose into R+L
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for (int i=1;i<n;++i)
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for (int j=1;j<m;++j)
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ans=min(ans,tilingRectangle(i,j)+solveL(n-i,m,n,m-j));
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// decompose into L+L
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for (int I=0;I<2;++I){
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for (int i=1;i<n;++i)
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for (int j=i+1;j<n;++j)
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for (int k=1;k<m;++k)
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ans=min(ans,solveL(i,m,j,k)+solveL(n-j,m,n-i,m-k));
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swap(n,m);
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}
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return rect[n][m] = ans;
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}
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int solveL(int n1, int m1, int n2, int m2){
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if (rect2[n1][m1][n2][m2])return rect2[n1][m1][n2][m2];
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return rect2[n1][m1][n2][m2] = min(Rect2(n1,m1,n2,m2),Rect2(m2,n2,m1,n1));
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}
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//L-shape
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int Rect2(int n1, int m1, int n2, int m2) //short1,long2,long1,short2
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{
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//恰好分为两个矩形(只需考虑一种,另一种已经在旋转中处理了)
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int ans = tilingRectangle(n1, m1 - m2) + tilingRectangle(n2, m2);
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//cut from a longer edge
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for (int i=1;i<n1;++i)
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ans=min(ans,tilingRectangle(i,m1)+solveL(n1-i,m1,n2-i,m2));
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//cut at another corner
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for (int i=1;i<n1;++i)
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ans=min(ans,tilingRectangle(i,m1-m2)+solveL(n1-i,m1,n2,m2));
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//枚举切分矩形的长度,注意旋转保证边长无负数
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for (int i = 1; i < m1; i++)
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if (m1 - i > m2)
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//注意L形的两种枚举
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ans = min(ans, tilingRectangle(n1, i) + solveL(n1, m1 - i, n2, m2));
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else if (m2 > m1 - i)
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ans = min(ans, tilingRectangle(n1, i) + solveL(n2 - n1, m2, n2, m1 - i));
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return ans;
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}
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}A;
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1. binary search. O(n^2 log U).
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2. dijkstra twice. proof: https://leetcode.cn/problems/modify-graph-edge-weights/solution/xiu-gai-tu-zhong-de-bian-quan-by-leetcod-66bg/
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O(1).
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closest pair. O(n).
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counting hamiltonian path. O(n^2*2^n).
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‎2701-2800/2742. Painting the Walls.txt

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knapsack. O(nU).
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Divide into cases. O(1).
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const int N=1005,CH=256,inf=1<<20;
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int f[CH],f1[CH];
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inline void upd(int &x,int y){if (y>x)x=y;}
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class Solution {
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public:
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int minimizeConcatenatedLength(vector<string>& a) {
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int n=a.size(),add=0,add1=0,ma=0,ma1=0;
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for (int i='a';i<='z';++i)f[i]=f1[i]=-inf;
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f[a[0][0]]=0; f1[a[0].back()]=0;
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for (int i=1;i<n;++i){
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char x=a[i-1][0],y=a[i-1].back(),x1=a[i][0],y1=a[i].back();
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int v=max(f[y1]+1+add,ma),v1=max(f1[x1]+1+add1,ma1);
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if (y==x1)++add,++ma;
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if (y1==x)++add1,++ma1;
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upd(ma1,v); upd(ma,v1);
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upd(f1[y],v-add1); upd(f[x],v1-add);
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}
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int ans=-max(ma,ma1);
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for (int i=0;i<n;++i)ans+=a[i].size();
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return ans;
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}
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};
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sorting, sliding window. O(sort(n)).
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enumerate the number of steps, which is O(log n). O(log n).
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