HZAU 1208 Color Circle

                                                           Color Circle

There are colorful flowers in the parterre in front of the door of college and form many
beautiful patterns. Now, you want to find a circle consist of flowers with same color. What
should be done ?
Assuming the flowers arranged as matrix in parterre, indicated by a N*M matrix. Every
point in the matrix indicates the color of a flower. We use the same uppercase letter to
represent the same kind of color. We think a sequence of points d1, d2, … dk makes up a
circle while:
1. Every point is different.
2. k >= 4
3. All points belong to the same color.
4. For 1 <= i <= k-1, di is adjacent to di+1 and dk is adjacent to d1. ( Point x is adjacent
to Point y while they have the common edge).
N, M <= 50. Judge if there is a circle in the given matrix.
Input
There are multiply test cases.
In each case, the first line are two integers n and m, the 2nd ~ n+1th lines is the given
n*m matrix. Input m characters in per line.
Output
Output your answer as “Yes” or ”No” in one line for each case.
Sample Input
3 3
AAA
ABA
AAA
Sample Output
Yes

爆搜！！！

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e2+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
 
///数组大小
 
char a[N][N],vis[N][N];
int n,m,ans;
int xx[4]={0,1,0,-1};
int yy[4]={1,0,-1,0};
int check(int x,int y)
{
    if(x<=0||x>n||y<=0||y>m)
        return 0;
    return 1;
}
void dfs(int x,int y,int dep)
{
    if(ans)return;
    for(int i=0;i<4;i++)
    {
        int xxx=x+xx[i];
        int yyy=y+yy[i];
        if(check(xxx,yyy)&&a[xxx][yyy]==a[x][y])
        {
            if(vis[xxx][yyy]&&dep-vis[xxx][yyy]+1>=4)
            {
                ans=1;
            }
            else if(!vis[xxx][yyy])
            {
                vis[xxx][yyy]=dep;
                dfs(xxx,yyy,dep+1);
                vis[xxx][yyy]=0;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        ans=0;
        for(int i=1;i<=n;i++)
        scanf("%s",a[i]+1);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                dfs(i,j,1);
                if(ans)break;
            }
            if(ans)break;
        }
        if(ans)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}