CodeForces 510B Fox And Two Dots（判断环的存在性，DFS一类题）

http://codeforces.com/problemset/problem/510/B

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:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. 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 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：

判断给定的地图中是否存在相同字母组成的环。

思路：

在 DFS 的回溯过程中，如果存在遍历时的情况是上一次已经遍历的，就存在这样的环，

难点在于怎么样去标记已经遍历的坐标，除了引入访问标记数组之外，还引入一个变量，

参考代码。

Code：

#include<cstdio>#include<cstring>const int MYDD=1103;int n,m;char now,map[64][64];int vis[64][64];int dx[]= {-1,1,0,0};int dy[]= {0,0,-1,1};bool DFS(int y,int x,int dd) {vis[y][x]=dd;for(int j=0; j<4; j++) {int gx=x+dx[j];int gy=y+dy[j];if(gx>=0&&gx<m&&gy>=0&&gy<n&&map[gy][gx]==now) {if(vis[gy][gx]==-1) {if(DFS(gy,gx,dd+1))return true;// 找到上一次 DFS 访问的坐标// 即一个坐标被访问了两次} else if(vis[gy][gx]!=dd-1)return true;}}return false;}int main() {scanf("%d%d",&n,&m);for(int j=0; j<n; j++)scanf("%s",map[j]);memset(vis,-1,sizeof(vis));//坐标初始化没有访问for(int j=0; j<n; j++) {for(int k=0; k<m; k++) {now=map[j][k];if(vis[j][k]==-1)if(DFS(j,k,0)) {puts("Yes");return 0;}}}puts("No");return 0;}/* By:Shyazhut */