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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:
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).
主要是用dfs判断有没有回到出发的点，并保证不走回头路（所以dfs函数里定义四个变量）
#include<cstdio> #include<cstring> int flag[55][55]; char str[55][55]; int n,m,k; int dx[4]={-1,1,0,0}; int dy[4]={0,0,-1,1}; void dfs(int x,int y,int cx,int cy) { if(flag[x][y] == 1) { k=1; return ; } flag[x][y]=1; int i,nx,ny; for(i = 0 ; i < 4 ; i++) { nx=x+dx[i]; ny=y+dy[i]; if(str[nx][ny] == str[x][y] && (nx != cx || ny != cy)) { dfs(nx,ny,x,y); } } } int main() { while(scanf("%d %d",&n,&m)!=EOF) { k=0; int i,j; memset(flag,0,sizeof(flag)); for(i = 1 ; i <= n ; i++) { scanf("%s",str[i]+1); } for(i = 1 ; i <= n ; i++) { for(j = 1 ; j <= m ; j++) { if(flag[i][j] == 1) continue; dfs(i,j,i,j); if(k == 1) break; } if(k == 1) break; } if(k == 1) printf("Yes\n"); else printf("No\n"); } }
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原文地址：http://www.cnblogs.com/yexiaozi/p/5770414.html
