HZAU1208——Color Circle（dfs）

Description
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 >= 43. 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

求图中是否有一个环路，其中的字母都相同
爆搜

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <math.h>#include <algorithm>#include <queue>#include <iomanip>#include <map>#include <cctype>#define INF 0x3f3f3f3f#define MAXN 1000005#define Mod 1000000007using namespace std;int n,m;char mp[55][55];int vis[55][55];int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};bool dfs(int x,int y,char tag,int step,int prex,int prey){    int tx,ty;    vis[x][y]=1;    if(step>=4)    {        for(int i=0;i<4;++i)        {            tx=x+dx[i];            ty=y+dy[i];            if(tx<1||tx>n||ty<1||ty>m)                continue;            if((tx!=prex||ty!=prey)&&mp[tx][ty]==tag&&vis[tx][ty])                return true;        }    }    for(int i=0;i<4;++i)    {        tx=x+dx[i];        ty=y+dy[i];        if(tx<1||tx>n||ty<1||ty>m)            continue;        if((tx!=prex||ty!=prey)&&mp[tx][ty]==tag&&!vis[tx][ty])        {            vis[tx][ty]=1;            if(dfs(tx,ty,tag,step+1,x,y))                return true;        }    }    return false;}int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i=1;i<=n;++i)            scanf("%s",mp[i]+1);        memset(vis,0,sizeof(vis));        int flag=0;        for(int i=1;i<=n;++i)        {            for(int j=1;j<=m;++j)            {                if(!vis[i][j])                {                    flag=dfs(i,j,mp[i][j],1,i,j);                    if(flag)                        break;                }            }            if(flag)                break;        }        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}