题目链接：
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of sizen×mcells, 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 dotsd1,d2,…,dkacycleif and only if it meets the following condition:
Thesekdots are different: ifi≠jthendiis different fromdj.kis at least 4.All dots belong to the same color.For all1≤i≤k-1:dianddi+1are adjacent. Also,dkandd1should also be adjacent. Cellsxandyare called adjacent if they share an edge.
Determine if there exists acycleon the field.
Input
The first line contains two integersnandm(2≤n,m≤50): the number of rows and columns of the board.
Thennlines follow, each line contains a string consisting ofmcharacters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output
Output "Yes" if there exists acycle, and "No" otherwise.
Sample test(s)
input
3 4AAAAABCAAAAA
output
Yes
input
3 4AAAAABCAAADA
output
No
input
4 4YYYRBYBYBBBYBBBY
output
Yes
input
7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB
output
Yes
input
2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ
output
No
Note
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：
求给出的矩阵中是否存在由相同字母围成的圈！
代码如下：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;char s[56][56];int vis[56][56];int n, m;int xx[4] = {0,-1,1,0};int yy[4] = {1,0,0,-1};char tt;int mark;int judge(int x, int y){if(x>=0 && x<n && y>=0 && y<m)return 1;elsereturn 0;}void dfs(int x, int y, int fx, int fy){if(!judge(x, y))return ;vis[x][y] = 1;for(int i = 0; i < 4; i++){int tx = x + xx[i];int ty = y + yy[i];if(judge(tx,ty) && s[x][y]==s[tx][ty] && (tx!=fx || ty!=fy)){if(vis[tx][ty]){mark = 1;return ;}dfs(tx, ty, x, y);}}return ;}int main(){while(~scanf("%d%d",&n,&m)){memset(vis,0,sizeof(vis));for(int i = 0; i < n ; i++){scanf("%s",s[i]);}mark = 0;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(!vis[i][j]){dfs(i,j,-1,-1);if(mark){printf("Yes\n");return 0;}}}}printf("No\n");}return 0;}
，快乐要懂得分享，才能加倍的快乐
