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ZOJ 2975 Kinds of Fuwas(暴力)

2016-02-19 17:44 218 查看
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+7

using 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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