fzu 2056 暴力

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题意：

现在有一个n*m的矩阵A，在A中找一个H*H的正方形，使得其面积最大且该正方形元素的和不大于 limit。

思路：水题

预处理矩形元素和，然后二分枚举最大边长，然后把这边长在整个矩形中试一遍（O(n*m)）看是否符合。总时间复杂度O(n*m*log(min(n,m))) 可暴

code：

#include<cstdio>#include<iostream>#include<sstream>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<queue>#include<map>#include<set>#include<cmath>#include<cctype>#include<cstdlib>using namespace std;#define INF 0x3f3f3f3f#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define mod 1000000007typedef pair<int,int> pii;typedef long long LL;//------------------------------const int maxn = 1005;int n,m,limit;int pre[maxn][maxn];int maze[maxn][maxn];void deal(){    memset(pre, 0, sizeof(pre));    for(int i = 1; i <= m; i++){        pre[1][i] = pre[1][i-1] + maze[1][i];    }    for(int i = 2; i <= n; i++){        int tmp = 0;        for(int j = 1; j <= m; j++){            tmp += maze[i][j];            pre[i][j] = pre[i-1][j] + tmp;        }    }}void init(){    scanf("%d%d%d",&n,&m,&limit);    for(int i = 1; i <= n; i++){        for(int j = 1; j <= m; j++){            scanf("%d",&maze[i][j]);        }    }    deal();}bool ok(int x){    for(int i = 1; i <= n-x+1; i++){        for(int j = 1; j <= m-x+1; j++){            if(pre[i+x-1][j+x-1] - pre[i-1][j+x-1] - pre[i+x-1][j-1] + pre[i-1][j-1] <= limit)                return true;        }    }    return false;}void solve(){    int lb = 0, ub = min(n,m);    while(lb <= ub){        int mid = (lb+ub) / 2;        if(ok(mid)) lb = mid+1;        else ub = mid-1;    }    printf("%d\n",ub * ub);}int main(){    int t;    scanf("%d",&t);    while(t--){        init();        solve();    }    return 0;}

我的二分出了点小问题，真实比了狗了...T_T

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