深度优先搜索练习之神奇的矩环

题目是让找一个长度不为一的环，用DFS即可。

#include <iostream>
#include <cstring>
using namespace std;
char mp[510][510];
bool vis[210][210];
int n, m;
int fx[] = {0, 0, 1, -1}, fy[] = {1, -1, 0, 0};
bool flag;

void dfs(int x, int y, int c, int d)
{
    vis[x][y] = 1;
    for(int i = 0; i < 4; i++)
    {
        int a = x + fx[i], b = y + fy[i];
        if(a <= n && b <= m && a >= 1 && b >= 1 && mp[a][b] == mp[x][y])
        {
            if(vis[a][b] == 0)//如果没有走过则继续向下走
                dfs(a, b, x, y);
            else if(a != c && b != d)//如果被走过看，如果a == c 且 b == d,则是长度为一的环
            {
                flag = 1;
                return;
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio(0);
    while(cin >> n >> m)
    {
        flag = 0;
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                cin >> mp[i][j];
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                if(vis[i][j] == 0)
                {
                    dfs(i, j, i, j);
                    if(flag)
                    break;
                }
            }
            if(flag)
            break;
        }
        if(flag)
        cout << "Yes\n";
        else
        cout << "No\n";
    }
    return 0;
}

标签：&&,210,int,练习,vis,flag,mp,矩环,神奇
来源： https://www.cnblogs.com/Flying-bullet/p/16230303.html