【FZU】Problem 2056 最大正方形

這題的關鍵就是預處理矩陣利用

dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i-1][j-1]

遞推出矩陣，使得我們以後每次已經利用O(1)的複雜度計算任意一塊矩陣

之後枚舉正方形左上角的座標二分邊長，時間複雜度爲n^2 log(n)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1005;
int n,m,t;
int mat[maxn][maxn];
int ans[maxn][maxn];
int get_value(int x1,int y1,int x2,int y2){
    return ans[x2][y2] - (ans[x1 - 1][y2] + ans[x2][y1 - 1] - ans[x1 - 1][y1 - 1]);
}
void solve(){
    int l = 1,r = min(n,m);
    int rans = 0;
    while(l <= r){
        int h = (l + r) >> 1;
        int ok = 0;
        for(int i = 1; i  + h - 1<= n; i++){
            for(int j = 1; j  + h - 1 <= m; j++){
                     int e = get_value(i,j,i + h - 1,j + h - 1);
                     if(e <= t){
                        ok = 1;
                        break;
                     }
            }
            if(ok) break;
        }
        if(ok){
            rans = h;
            l = h +1;
        }
        else
            r = h - 1;
    }
    printf("%d\n",rans * rans);
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(ans,0,sizeof(ans));
        scanf("%d%d%d",&n,&m,&t);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                scanf("%d",&mat[i][j]);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                ans[i][j] = ans[i - 1][j] + ans[i][j - 1] - ans[i - 1][j - 1] + mat[i][j];
        solve();
    }
    return 0;
}