2016CCPC东北地区大学生程序设计竞赛 1005 HDU5926

链接http://acm.hdu.edu.cn/showproblem.php?pid=5926

题意：给我们一个矩阵，问你根据连连看的玩法可以消去其中的元素

解法：连连看怎么玩，就怎么写，别忘记边界

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int t;
int a[100][100];
int main()
{
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        int flag=0;
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>a[i][j];
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(i+1<=n&&a[i+1][j]==a[i][j])
                {
                    flag=1;
                    break;
                }
                if(j+1<=n&&a[i][j]==a[i][j+1])
                {
                   // cout<<"A"<<endl;
                    flag=1;
                    break;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(i!=j)
                {
                    if(a[1][i]==a[1][j]||a[n][i]==a[n][j])
                    {
                    //    cout<<a[0][i]<<" "<<a[0][j]<<endl;
                      //  cout<<"A"<<endl;
                        flag=1;
                        break;
                    }
                    if(a[i][1]==a[j][1]||a[i][m]==a[j][m])
                    {
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(flag)
        {
            printf("Case #%d: Yes\n",i );
        }
        else
        {
             printf("Case #%d: No\n",i );
        }
    }
    return 0;
}