D - Fox And Two Dots
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.
InputThe 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).
代码：
#include<cstdio>
#include <cstring>
int m,n,a[55][55];
char map[55][55];
int x_move[4]={0,0,-1,1};
int y_move[4]={1,-1,0,0};
int flag,fx,fy;
void dfs(int w,int h,int fx,int fy,char s)//fx,fy在这里记录父节点 ，w,h本次开始搜索的位置
{if(flag==1) //成环则结束，返回 return ;for(int i=0;i<4;i++){int nx=w+x_move[i],ny=h+y_move[i];if(nx>=0&&ny>=0&&nx<m&&ny<n&&map[nx][ny]==s){if(nx==fx&&ny==fy)//如果搜索到的点和父节点相同，则跳过 continue;if(a[nx][ny]==1)//如果搜索到之前已经搜索到过的点，说明成环，标记 {flag=1;return ;} a[nx][ny]=1; //搜索过的做标记dfs(nx,ny,w,h,s); //让当前点做下一点的父节点 }}
}
int main()
{while(~scanf("%d%d",&m,&n)){for(int i=0;i<m;i++){scanf("%s",map[i]);}memset(a,0,sizeof(a)); //将数组a[i][j]全部初始化成0flag=0;fx=-5,fy=-5; //初始化父节点，fx,fy等于-5时一定是搜不到的 for(int i=0;i<m;i++){for(int j=0;j<n;j++){if(a[i][j]==0) {a[i][j]=1; //将搜索过的做标记 dfs(i,j,fx,fy,map[i][j]);if(flag==1)break;}}if(flag==1)break;}if(flag==1) printf("Yes\n");else printf("No\n");}return 0;
}