ZOJ 2975Kinds of Fuwas（暴力）

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题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2975

给出n*m的矩阵，要求算出四个角类型相同的小矩形的数目
使用暴力枚举任意两列，然后枚举每一行，统计每行的两列相同的个数，然后使用组合数求解，时间复杂度250^3.

#include<iostream>#include<cstdio>#include<set>#include<string>#include<string.h>#include<cstring>#include<vector>#include<map>#include<queue>#include<stack>#include<cctype>#include<algorithm>#include<sstream>#define mt(a) memset(a,0,sizeof a)#define fl(a,b,c) fill(a,b,c)#define inf 1000000000+7using namespace std;typedef long long ll;char graph[300][300];ll g[5];int main(){    int T;    cin >> T;    while (T--)    {        int n, m;;        ll ans = 0;        memset(graph, 0, sizeof graph);        memset(g, 0, sizeof g);        scanf("%d %d", &n, &m);        for (int i = 0; i < n; i++)            scanf("%s", graph[i]);        for (int i = 0; i < m; i++)        {            for (int j = i + 1; j < m; j++)            {                memset(g, 0, sizeof g);                for (int k = 0; k < n; k++)                {                    if (graph[k][i] == graph[k][j])                    {                        if (graph[k][i] == 'B')g[0]++;                        else if (graph[k][i] == 'J')g[1]++;                        else if (graph[k][i] == 'H')g[2]++;                        else if (graph[k][i] == 'Y')g[3]++;                        else if (graph[k][i] == 'N')g[4]++;                    }                }                for (int k = 0; k < 5; k++)                {                    if(g[k])ans += ((g[k] * (g[k] - 1)) / 2);                }            }        }        printf("%lld\n", ans);    }    return 0;}

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