Codeforces 510B Fox And Two Dots 【DFS】

好久好久，都没有写过搜索了，看了下最近在CF上有一道DFS水题 = =

数据量很小，爆搜一下也可以过

额外注意的就是防止往回搜索需要做一个判断。

Source code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <bits/stdc++.h>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 1              ;
const int M = 200000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;

char mp [80][80];
int n,m;
int Dir [4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
int Vis [80][80];

int Fit(int x , int y){
    return x >= 0 && x < n && y >= 0 && y < m;
}

int Dfs (int x, int y, int px, int py, char c){
    Vis[x][y] = 1;
    for(int i = 0 ; i < 4 ; ++i){
        int dx = x + Dir[i][0];
        int dy = y + Dir[i][1];
        if(dx == px && dy == py)    continue;
        if (Fit (dx , dy) && mp[dx][dy] == c){
            if (Vis[dx][dy]){
                return 1;
            }
            if (Dfs (dx ,dy, x, y, mp[dx][dy])){
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    int i ,j ;
    while(cin >> n >> m){
        memset(Vis, 0, sizeof(Vis));
        for(i = 0; i < n; ++i){
            for(j = 0; j < m; ++j){
                cin >> mp[i][j];
            }
        }
        for(i = 0; i < n; ++i){
            for(j = 0; j < m; ++j){
                if(!Vis[i][j]){
                    if(Dfs(i, j, -1, -1, mp[i][j])){
                        cout << "Yes" << endl;
                        return 0;
                    }
                }
            }
        }
        cout << "No" << endl;
    }
    return 0;
}