Fox And Two Dots

Fox And Two Dots Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 510B Appoint description: Description 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. Input 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 Output "Yes" if there exists a cycle, and "No" otherwise. Sample Input Input 3 4 AAAA ABCA AAAA Output Yes Input 3 4 AAAA ABCA AADA Output No Input 4 4 YYYR BYBY BBBY BBBY Output Yes Input 7 6 AAAAAB ABBBAB ABAAAB ABABBB ABAAAB ABBBAB AAAAAB Output Yes Input 2 13 ABCDEFGHIJKLM NOPQRSTUVWXYZ Output No Hint 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).

题意就是，任意以一个字母为开头，然后沿着与这个字母相同的字母走（不能回头走）如果能转一圈回到最开始那个字母，就输出Yes，否则输出No；实际还是dfs，只要加个限制条件防止回头走就行了；

代码中的fx，fy是为了防止走回头路：

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; bool vis[55][55]; char map[55][55]; int dx[4]={1,0,0,-1}; int dy[4]={0,-1,1,0}; int ex,ey,flag; int n,m; void dfs(int x,int y,int fx,int fy,char s) { if(flag==1){ return ; } for(int i=0;i<4;i++) { ex=x+dx[i]; ey=y+dy[i]; if(ex>=0&&ey>=0&&ex<n&&ey<m&&map[ex][ey]==s) { if(ex==fx&&ey==fy){ continue; } else{ if(vis[ex][ey]&&map[ex][ey]==s){ flag=1; return ; } } vis[ex][ey]=true; dfs(ex,ey,x,y,map[ex][ey]); } } } int main() { flag=0; memset(vis,false,sizeof(vis)); scanf("%d%d",&n,&m); for(int i=0;i<n;i++){ scanf("%s",map[i]); } for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(!vis[i][j]) { vis[i][j]=true; dfs(i,j,-1,-1,map[i][j]); if(flag==1) break; } } if(flag==1) break; } if(flag==1) printf("Yes\n"); else printf("No\n"); return 0; }