codeforces-Fox And Two Dots【DFS】（思维）

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B. Fox And Two Dots time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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.

Examples 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 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：在n*m的地图上，每个格子涂有不同的颜色（A~Z），问是否有某一种颜色能组成一个回路且长度至少为4. 

#include<cstdio> #include<algorithm> #include<cstring> using namespace std; int n,m; char map[110][110]; bool vis[110][110]; int dx[4]={-1,1,0,0}; int dy[4]={0,0,-1,1}; bool judge(int x,int y) { if(x>=0&&x<n&&y>=0&&y<m) return 1; return 0; } bool DFS(int x,int y,int px,int py,int cnt) { vis[x][y]=1; for(int i=0;i<4;i++) { int nx1=x+dx[i]; int ny1=y+dy[i]; if(vis[nx1][ny1]&&judge(nx1,ny1)&&(nx1!=px||ny1!=py)&&cnt>3) // 里面第三个判断条件是保证不会回头计算上一步刚走过的那个点 return 1; } char c=map[x][y]; for(int i=0;i<4;i++) { int nx2=x+dx[i]; int ny2=y+dy[i]; if(!vis[nx2][ny2]&&judge(nx2,ny2)&&map[nx2][ny2]==c) { return DFS(nx2,ny2,x,y,cnt+1); // 这个不是 void 型的 DFS 所以调用自己要加 return } } return 0; } int main() { while(~scanf("%d %d",&n,&m)) { bool flag=0; 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++) { memset(vis,0,sizeof(vis)); if( DFS(i,j,i,j,1) ) { flag=1; break; } } if(flag) break; } if(flag) printf("Yes\n"); else printf("No\n"); } return 0; }