B. Fox And Two Dots
time limit per test
memory limit per test
input
output
n × m
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
d1, d2, ..., dk a cycle
- k dots are different: if i ≠ j then di is different from dj.
- k
- All dots belong to the same color.
- 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y
cycle
Input
n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
n lines follow, each line contains a string consisting of m
Output
Yes" if there exists a cycle, and "No" otherwise.
Sample test(s)
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
A' form a cycle.
In second sample there is no such cycle.
Y' = Yellow, 'B' = Blue, 'R' = Red).
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
int n,m;
char map[51][51];
int v[51][51];
int jx[] = {0,1,0,-1};
int jy[] = {1,0,-1,0};
int flag = 0;
struct node
{
int x;
int y;
};
void DFS(int xx,int yy,int x1,int y1)
{
for(int i=0; i<4; i++)
{
if(flag == 1)
{
return ;
}
int x,y;
x = xx + jx[i];
y = yy + jy[i];
if(v[x][y] == 1 && map[xx][yy] == map[x][y] && x1!=x && y1!=y && x>=0 && x<n && y>=0 && y<m)
{
printf("Yes\n");
flag = 1;
return ;
}
if(v[x][y] == 0)
{
if(map[xx][yy] == map[x][y] && x>=0 && x<n && y>=0 && y<m)
{
//printf("x = %d y = %d\n",x,y);
v[x][y] = 1;
DFS(x,y,xx,yy);
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
flag = 0;
memset(v,0,sizeof(v));
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++)
{
if(v[i][j] == 0 && flag == 0)
{
DFS(i,j,i,j);
}
}
}
if(flag == 0)
{
printf("No\n");
}
}
return 0;
}
