HDU 5926 Mr. Frog's Game 【模拟】 （2016CCPC东北地区大学生程序设计竞赛）

Mr. Frog’s Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 277    Accepted Submission(s): 192

Problem Description

One day, Mr. Frog is playing Link Game (Lian Lian Kan in Chinese).

In this game, if you can draw at most three horizontal or vertical head-and-tail-connected lines over the empty grids(the lines can be out of the whole board) to connect two non-empty grids with the same symbol or the two non-empty grids with the same symbol are adjacent, then you can change these two grids into empty and get several more seconds to continue the game.

Now, Mr. Frog starts a new game (that means there is no empty grid in the board). If there are no pair of grids that can be removed together，Mr. Frog will say ”I’m angry” and criticize you.

Mr. Frog is battle-scarred and has seen many things, so he can check the board in a very short time, maybe one second. As a Hong Kong Journalist, what you should do is to check the board more quickly than him, and then you can get out of the room before Mr. Frog being angry.

 

Input

The first line contains only one integer T (T≤500), which indicates the number of test cases.

For each test case, the first line contains two integers n and m (1≤n,m≤30).

In the next n lines, each line contains m integers,  j-th number in the i-th line means the symbol on the grid(the same number means the same symbol on the grid).

 

Output

For each test case, there should be one line in the output.

You should output “Case #x: y”，where x is the case number(starting from 1), and y is a string representing the answer of the question. If there are at least one pair of grids that can be removed together, the y is “Yes”（without quote), else y is “No”.

 

Sample Input

2
3 3
1 2 1
2 1 2
1 2 1
3 3
1 2 3
2 1 2
3 2 1

 

Sample Output

Case #1: Yes
Case #2: No

Hint

first sample can be explained as below.

 

Source

2016CCPC东北地区大学生程序设计竞赛 - 重现赛

 

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题目链接：

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

题目大意：

　　N*M(N,M<=30)的格子，求连连看第一步是否能连。

题目思路：

　　【模拟】

　　+1s 直接枚举相邻的，最外层的能否跨接即可。

 //
 //by coolxxx
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<algorithm>
 #include<string>
 #include<iomanip>
 #include<map>
 #include<stack>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<memory.h>
 #include<time.h>
 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 //#include<stdbool.h>
 #include<math.h>
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 #define min(a,b) ((a)<(b)?(a):(b))
 #define max(a,b) ((a)>(b)?(a):(b))
 #define abs(a) ((a)>0?(a):(-(a)))
 #define lowbit(a) (a&(-a))
 #define sqr(a) ((a)*(a))
 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 #define mem(a,b) memset(a,b,sizeof(a))
 #define eps (1e-10)
 #define J 10000
 #define mod 1000000007
 #define MAX 0x7f7f7f7f
 #define PI 3.14159265358979323
 #define N 34
 using namespace std;
 typedef long long LL;
 double anss;
 LL aans;
 int cas,cass;
 int n,m,lll,ans;
 int a[N][N];
 int dx[]={-,,,};
 int dy[]={,,-,};
 bool judge()
 {
     int i,j,k;
     for(i=;i<n;i++)
         for(j=;j<m;j++)
             for(k=;k<;k++)
                 if(a[i][j]==a[i+dx[k]][j+dy[k]])return ;
     for(i=;i<n;i++)
         for(j=i+;j<=n;j++)
             if(a[][i]==a[][j] || a[n][i]==a[n][j])return ;
     for(i=;i<m;i++)
         for(j=i+;j<=m;j++)
             if(a[i][]==a[j][] || a[i][m]==a[j][m])return ;
     return ;
 }
 int main()
 {
     #ifndef ONLINE_JUDGEW
 //    freopen("1.txt","r",stdin);
 //    freopen("2.txt","w",stdout);
     #endif
     int i,j,k;
     int x,y,z;
 //    init();
 //    for(scanf("%d",&cass);cass;cass--)
     for(scanf("%d",&cas),cass=;cass<=cas;cass++)
 //    while(~scanf("%s",s))
 //    while(~scanf("%d%d",&n,&m))
     {
         printf("Case #%d: ",cass);
         scanf("%d%d",&n,&m);
         for(i=;i<=n;i++)
             for(j=;j<=m;j++)
                 scanf("%d",&a[i][j]);
         if(judge())puts("Yes");
         else puts("No");
     }
     return ;
 }
 /*
 //

 //
 */