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.
Sample 1
| Inputcopy | Outputcopy |
|---|---|
3 4 AAAA ABCA AAAA | Yes |
Sample 2
| Inputcopy | Outputcopy |
|---|---|
3 4 AAAA ABCA AADA | No |
Sample 3
| Inputcopy | Outputcopy |
|---|---|
4 4 YYYR BYBY BBBY BBBY | Yes |
Sample 4
| Inputcopy | Outputcopy |
|---|---|
7 6 AAAAAB ABBBAB ABAAAB ABABBB ABAAAB ABBBAB AAAAAB | Yes |
Sample 5
| Inputcopy | Outputcopy |
|---|---|
2 13 ABCDEFGHIJKLM NOPQRSTUVWXYZ | 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).
#include <iostream>
using namespace std;
#include<string>
#include<vector>
#include<algorithm>
#include<string.h>
#pragma warning (disable:4996)
#include<stack>
#include<set>
bool flag = false;
int n, m;
int book[55][55];
char graph[55][55];
int diry[4] = { 1,-1,0,0 };
int dirx[4] = { 0,0,1,-1 };
void dfs(int x, int y, int frox, int fory, char ch);
int main() {
cin >> n >> m;
for (int cnt = 0; cnt < n; cnt++)
for (int cnt1 = 0; cnt1 < m; cnt1++)
cin >> graph[cnt][cnt1];
for (int cnt = 0; cnt < n; cnt++) {
for (int cnt1 = 0; cnt1 < m; cnt1++) {
char temp = graph[cnt][cnt1];
if (!book[cnt][cnt1]){
dfs(cnt1, cnt, -1, -1, temp);//cnt是行数,cnt1是列数
if (flag) {
cout << "Yes" << endl;
return 0;
}
}
}
}
cout << "No" << endl;
return 0;
}
void dfs(int x, int y, int frox, int froy, char ch) {
if (x<0 || y<0 || x>m - 1 || y>n - 1)//n是行数,m是列数
return;
if (ch != graph[y][x])
return;
if (book[y][x]) {
flag = true;
return;
}
book[y][x] = true;
for (int cnt = 0; cnt < 4; cnt++) {
int nex = dirx[cnt] + x;
int ney = diry[cnt] + y;
if (nex == frox && ney == froy)
continue;
dfs(nex, ney, x, y, ch);
}
}
思路:就是通过深度搜索,只要遍到原点就可以了;
我踩过的坑,我一直卡测试点11,查了好几遍,发现是因为x,y在输入进函数时写反了,然后猜测可能是因为这样会导致某个节点没有扫到,或者说越界扫了某个节点,然后本来那个节点是满足题意的.
