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;}