codeforces 510BFox And Two Dots 深搜dfs
Fox And Two Dots
CodeForces - 510B
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.
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 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
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).
思路：
并查集写法：
对于每个字母，我们都赋给他们一个不同的值，然后按行从左往右扫，对于扫到的元素，我们找到它本身和它向左和向上的根节点（如果这三个点的字母相同的话），找这两个的作用是一个用来与它建立一个集合，另一个就可以用它过来判环了，于是这道题就转变成了无想吐的判环问题
dfs写法：
对于相同的字母我们进行搜索，但是要记录一下它上次走过的位置，这样只要从当前状态走到了一个走过的点，他就一定是成环的
代码：
并查集：
#include
using namespace std;
#define N 55
char ch[N][N];
int num[N][N],f[N * N],n,m;
int dx[2] = {-1,0};
int dy[2] = {0,-1};
bool check(int x,int y)
{
if (x >= 0 && x < n && y >= 0 && y < m)
return 1;
return 0;
}
int find(int x)
{
if (f[x] != x)
f[x] = find(f[x]);
return f[x];
}
int main()
{
int pos = 1;
scanf("%d %d",&n,&m);
for (int i = 1;i <= n * m;i ++) f[i] = i;
for (int i = 0;i < n;i ++) scanf("%s",ch[i]);
for (int i = 0;i < n;i ++)
for (int j = 0;j < m;j ++)
{
num[i][j] = pos ++;
for (int k = 0;k < 2;k ++)
{
int ex = i + dx[k],ey = j + dy[k];
if (check(ex,ey) && ch[ex][ey] == ch[i][j])
{
int u = find(num[i][j]),v = find(num[ex][ey]);
if (u == v)
{
puts("Yes");
return 0;
}
f[u] = v;
}
}
}
puts("No");
return 0;
} dfs:
#include
using namespace std;
#define N 55
int n,m;
char mp[N][N];
bool vis[N][N],flag = 0;
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
bool check(int x,int y)
{
if (x >= 0 && x < n && y >= 0 && y < m)
return 1;
return 0;
}
void dfs(int x,int y,int lx,int ly,char w)
{
vis[x][y] = 1;
for (int i = 0;i < 4;i ++)
{
int ex = x + dx[i],ey = y + dy[i];
if (ex == lx && ey == ly) continue;
if (check(ex,ey) && mp[ex][ey] == w)
{
if (vis[ex][ey])
{
flag = 1;
break;
}
dfs(ex,ey,x,y,w);
}
}
}
int main()
{
scanf("%d %d",&n,&m);
for (int i = 0 ;i < n;i ++)
scanf("%s",mp[i]);
for (int i = 0;i < n;i ++)
for (int j = 0;j < m;j ++)
{
if (!vis[i][j])
{
dfs(i,j,-1,-1,mp[i][j]);
if (flag)
{
puts("Yes");
return 0;
}
}
}
puts("No");
return 0;
}