Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n?×?m cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1,?d2,?...,?dk a cycle if and only if it meets the following condition:
- These k dots are different: if i?≠?j then di is different from dj.
- k is at least 4.
- All dots belong to the same color.
- For all 1?≤?i?≤?k?-?1: di and di?+?1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
The first line contains two integers n and m (2?≤?n,?m?≤?50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output "Yes" if there exists a cycle, and "No" otherwise.
3 4 AAAA ABCA AAAA
Yes
3 4 AAAA ABCA AADA
No
4 4 YYYR BYBY BBBY BBBY
Yes
7 6 AAAAAB ABBBAB ABAAAB ABABBB ABAAAB ABBBAB AAAAAB
Yes
2 13 ABCDEFGHIJKLM NOPQRSTUVWXYZ
No
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
题意:给你N*M的字符矩阵问是否在矩阵中能够找到一个由同种字符围成的环
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=55;
char map[maxn][maxn];
bool vis[maxn][maxn];
int n,m;
int mov[][2]={0,1,0,-1,1,0,-1,0};
bool judge(int x,int y){return x>=0&&x<n&&y>=0&&y<m;
}
bool sign=false;
void dfs(int x,int y,char color,int dir){//printf("%d %d %c\n",x,y,color);if(sign)return ;if(vis[x][y]){//printf("x=%d y=%d\n",x,y);sign=true;return ;}for(int i=0;i<4;++i){if(dir==0&&i==1)continue;if(dir==1&&i==0)continue;if(dir==2&&i==3)continue;if(dir==3&&i==2)continue;int xx=x+mov[i][0];int yy=y+mov[i][1];if(judge(xx,yy)&&color==map[xx][yy]){vis[x][y]=true;dfs(xx,yy,color,i);vis[x][y]=false;}}
}
int main()
{int i,j;scanf("%d%d",&n,&m);for(i=0;i<n;++i){scanf("%s",map[i]);}for(i=0;i<n;++i){for(j=0;j<m;++j){//if(vis[i][j])continue;dfs(i,j,map[i][j],-1);if(sign)break;//printf("%d ",i,j);}if(sign)break;}if(sign){printf("Yes\n");}else {printf("No\n");}return 0;
}