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.
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.
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).
此题大意：
给出一个图形，问图形中是否存在一个环；
分析：
从一个点出发形成一个环后回到出发点，所需步骤最少为4，以此为判断条件，运用dfs求解；
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; char map1[51][51]; int visit[51][51]; int dir[][2]={{1,0},{-1,0},{0,1},{0,-1}}; int n,m,step; int dfs(char c,int sx,int sy,int cx,int cy) { if(sx==cx&&sy==cy&&visit[cx][cy]) { return step>=4?1:0; } int i; visit[cx][cy]=1;//标记已走 for(i=0;i<4;i++) { int nx=cx+dir[i][0]; int ny=cy+dir[i][1]; if((nx>=0&&nx<n&&ny>=0&&ny<m&&map1[nx][ny]==c&&!visit[nx][ny])||(nx==sx&&ny==sy)) { step++; if(dfs(c,sx,sy,nx,ny)) return 1; step--; } } return 0; } int slove() { int i,j; for(i=0;i<n;i++) { for(j=0;j<m;j++) { memset(visit,0,sizeof(visit));//对所有点进行遍历 step=0;//记录步数 if(dfs(map1[i][j],i,j,i,j)) return 1; } } return 0; } int main() { int i; while(~scanf("%d%d",&n,&m)) { for(i=0; i<n; i++) scanf("%s",map1[i]);//图形输入； int ans=slove(); printf("%s\n",ans?"Yes":"No"); } return 0; }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.
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.
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).
