USACO3.3.4 Home on the Range (range)

dp[i][j]：以点(i,j)为右下角能取到的最大正方形的边长

若arr[i][j]==0,则dp[i][j]=0

否则dp[i][j]=min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1

/*
ID:shijiey1
PROG:range
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n;
int arr[258][258];
int dp[258][258];
int ans[258];
int main() {
freopen("range.in", "r", stdin);
freopen("range.out", "w", stdout);
scanf("%d\n", &n);
char c;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
scanf("%c", &c);
arr[i][j] = c - '0';
}
getchar();
}

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (!arr[i][j]) dp[i][j] = 0;
else dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
if (dp[i][j] > 1);
ans[dp[i][j]]++;
}
}
for (int i = 2; i <= n; i++) {
int re = 0;
for (int j = i; j <= n; j++) {
re += ans[j];
}
if (re)
printf("%d %d\n", i, re);
}
return 0;
}