HDU 5706 GirlCat (DFS，暴力)

题意：给定一个n*m的矩阵，然后问你里面存在“girl”和“cat”的数量。

析：很简单么，就是普通搜索DFS，很少量。只要每一个字符对上就好，否则就结束。

代码如下：

#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1000 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
char s[maxn][maxn];
int n, m;
int vis[maxn][maxn];

inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int dfs(int r, int c, char *t, int x, bool ok){
    if(x == 3 && ok){
        if(s[r][c] == t[x])  return 1;
        return 0;
    }
    else if(!ok && x == 2){
        if(s[r][c] == t[x])  return 1;
        return 0;
    }

    int ans = 0;
    for(int i = 0; i < 4; ++i){
        int xx = r + dr[i];
        int yy = c + dc[i];
        if(is_in(xx, yy) && s[xx][yy] == t[x+1])  ans += dfs(xx, yy, t, x+1, ok);
    }
    return ans;
}

int main(){
    int T; cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 0; i < n; ++i)
            scanf("%s", s[i]);

        int ans1 = 0;
        int ans2 = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(s[i][j] == ‘g‘){
                    ans1 += dfs(i, j, "girl", 0, true);
                }
                else if(s[i][j] == ‘c‘)
                    ans2 += dfs(i, j, "cat", 0, false);
            }
        }
        cout << ans1 << " " << ans2 << endl;
    }
    return 0;
}