标签：codeforces dfs
题目链接：http://codeforces.com/problemset/problem/510/B
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:
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.
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).
题意：
求给出的矩阵中是否存在由相同字母围成的圈！
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
char s[56][56];
int vis[56][56];
int n, m;
int xx[4] = {0,-1,1,0};
int yy[4] = {1,0,0,-1};
char tt;
int mark;
int judge(int x, int y)
{
if(x>=0 && x<n && y>=0 && y<m)
return 1;
else
return 0;
}
void dfs(int x, int y, int fx, int fy)
{
if(!judge(x, y))
return ;
vis[x][y] = 1;
for(int i = 0; i < 4; i++)
{
int tx = x + xx[i];
int ty = y + yy[i];
if(judge(tx,ty) && s[x][y]==s[tx][ty] && (tx!=fx || ty!=fy))
{
if(vis[tx][ty])
{
mark = 1;
return ;
}
dfs(tx, ty, x, y);
}
}
return ;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(vis,0,sizeof(vis));
for(int i = 0; i < n ; i++)
{
scanf("%s",s[i]);
}
mark = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if(!vis[i][j])
{
dfs(i,j,-1,-1);
if(mark)
{
printf("Yes\n");
return 0;
}
}
}
}
printf("No\n");
}
return 0;
}
CodeForces 510 B. Fox And Two Dots（DFS 啊）
标签：codeforces dfs
原文地址：http://blog.csdn.net/u012860063/article/details/43458211
