Largest Square

题目链接: link.
Given a matrix (H × W) which contains only 1 and 0, find the area of the largest square matrix which only contains 0s.

Input
H W
c1,1 c1,2 … c1,W
c2,1 c2,2 … c2,W
:
cH,1 cH,2 … cH,W
In the first line, two integers H and W separated by a space character are given. In the following H lines, ci,j, elements of the H × W matrix, are given.

Output
Print the area (the number of 0s) of the largest square.

Constraints
1 ≤ H, W ≤ 1,400
Sample Input
4 5
0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 0 0 1 0
Sample Output
4
题意： 给一个H×W个边长为1的正方形瓷砖排列在一起，0代表瓷砖干净，1代表瓷砖有污渍，输出仅有干净瓷砖构成的最大正方形的面积。
思路： 用dp(i,j)来表示从瓷砖(i,j)向左上方扩展可形成的最大正方形的边长（瓷砖数），那么要知道dp(i,j)就要知道dp(i-1,j)、dp(i,j-1)和dp(i-1,j-1)三个值，也就是左上方、上方、左侧的最大正方形的值。
因为dp[i][j]以右下角向左上角扩展形成最大正方形，能向外拓展的最大长度，取决于那条边最先碰到1（有污渍的瓷砖），而这就需要 d p ( i − 1 , j ) dp(i-1,j) dp(i−1,j)(左延伸)、 d p ( i , j − 1 ) dp(i,j-1) dp(i,j−1)(上延伸)和dp ( i − 1 , j − 1 ) (i-1,j-1) (i−1,j−1)(斜着延伸)算出来的值，来取最小值，最小值，代表最先碰到有污渍的瓷砖，一碰有污渍的瓷砖就无法再往外拓展了。
当前dp[i][j]如果是有污渍的瓷砖，就无法形成正方形，直接赋值为0即可。
(明明是找最大正方形，方程确实最小值，着实有点反直觉 ）
综上所述，动态转移方程就是：
d p [ i ] [ j ] = m i n ( d p [ i − 1 ] [ j − 1 ] , m i n ( d p [ i − 1 ] [ j ] , d p [ i ] [ j − 1 ] ) ) + 1 dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1 dp[i][j]=min(dp[i−1][j−1],min(dp[i−1][j],dp[i][j−1]))+1

      //动态转移的核心
for(int i=1;i<H;i++) {for(int j=1;j<W;j++) {if(g[i][j]==1) { dp[i][j]=0;}else {dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;res=max(res,dp[i][j]); //找最大边长（最大面积）}}}

#include <algorithm>
#include <iostream>
using namespace std;
const int N = 1411;int dp[N][N], g[N][N];int solve(int H, int W) {int res = 0;for (int i = 0; i < H; i++) {for (int j = 0; j < W; j++) {dp[i][j] = (g[i][j] + 1) % 2;res |= dp[i][j];}}for (int i = 1; i < H; i++) {for (int j = 1; j < W; j++) {if (g[i][j] == 1) {dp[i][j] = 0;} else {dp[i][j] =min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;res = max(res, dp[i][j]);}}}return res * res;
}
int main() {int H, W;scanf("%d%d", &H, &W);for (int i = 0; i < H; i++) {for (int j = 0; j < W; j++) {scanf("%d", &g[i][j]);}}printf("%d\n", solve(H, W));return 0;
}

To be continued
如果你有任何建议或者批评和补充，请留言指出，不胜感激

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