ZOJ 2975 Kinds of Fuwas

枚举。

枚举两行，然后算这两行之间有多少个矩形满足条件。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
}

int T,n,m;
char s[300][300];

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++) scanf("%s",s[i]);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                int B=0,J=0,H=0,Y=0,N=0;
                for(int k=0;k<m;k++)
                {
                    if(s[i][k]=='B'&&s[j][k]=='B') B++;
                    if(s[i][k]=='J'&&s[j][k]=='J') J++;
                    if(s[i][k]=='H'&&s[j][k]=='H') H++;
                    if(s[i][k]=='Y'&&s[j][k]=='Y') Y++;
                    if(s[i][k]=='N'&&s[j][k]=='N') N++;
                }
                ans=ans+B*(B-1)/2+J*(J-1)/2+H*(H-1)/2+Y*(Y-1)/2+N*(N-1)/2;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}