HDU 2830 Matrix Swapping II

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这个题好像让我钻空子了。也是求完全是 1的子矩阵的 1的最多的个数。

方法还是对于每一个是 1 的位置。先求出竖直方向连接的1的个数，然后对于每一行非 0的排序乘就好了。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <fstream>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <vector>#include <cmath>#include <iomanip>#include <cmath>typedef long long LL;typedef unsigned long long LLU;const double PI=acos(-1.0);using namespace std;#define MAXN 1000+10#define INF 1 << 30char s[MAXN][MAXN],d[MAXN][MAXN];int q[MAXN][MAXN];int main (){    int m,n;    while(scanf("%d%d",&m,&n) != EOF){        for(int i = 1; i <= m; i++){            scanf("%s",s[i]+1);        }        memset(q,0,sizeof(q));        memset(d,0,sizeof(d));        for(int i = 1; i <= m; i++){            for(int j = 1; j <= n; j++){                if(s[i][j] == '0')                    q[i][j] = 0;                else                    q[i][j] = 1;            }        }        for(int i = 1; i <= m; i++){            for(int j = 1; j <= n; j++){                if(q[i][j])                    d[i][j] = d[i-1][j] + 1;                else                    d[i][j] = 0;            }        }        int ans = 0;        for(int i = 1; i<= m; i++){            vector <int> _d;            for(int j = 1; j <= n; j++)                if(d[i][j])                _d.push_back(d[i][j]);            sort(_d.begin(),_d.end());            int len = _d.size();            for(int j = 0; j < len; j++){                ans = max(ans, _d[j]*(len-j));            }        }        printf("%d\n",ans);    }    return 0;}

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