CodeForces

http://codeforces.com/problemset/problem/586/D

题意：有一个3*n（n<100）的隧道。一个人在最左边，要走到最右边，每次他先向右移动一格，再上下移动一格。隧道里有车，每次人移动以后，车会向右移动两格，人与车轮流运动。

题解，将每次车的移动变成人往右移动两格，然后就能搜索了。dfs，用vis数组判掉已经移动到的格子（因为已经移动到的格子要么是不能走到底的，要么就可以走完）。注意他只能移动到三的倍数的格子，所以读入map后要再后面加两列'.'

#define  _CRT_SECURE_NO_WARNINGS
#include<cstdio>  
#include<algorithm>  
#include<iostream>
#include<string>
#include <vector>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn = 100 + 5;
//int d[maxn], k[maxn];
string map[4];
int vis[4][maxn];
//1e4
int n, x;
int ok;

void  dfs(int dx, int dy) {
    if (vis[dx][dy]) return;
    vis[dx][dy] = 1;
    for (int i = -1; i <= 1; i++) {
        int ddx = dx + i;
        if (dy >= n - 1) {
            ok = 1;
            return;
        }
        if (map[dx][dy + 1] != '.')return;
        if (map[dx][dy + 2] == '.'&&map[dx][dy + 3] == '.')dfs(dx, dy + 3); if (ok)return;
        if (dx - 1 > 0 && map[dx - 1][dy + 1] == '.'&&map[dx - 1][dy + 2] == '.'&&map[dx - 1][dy + 3] == '.') dfs(dx - 1, dy + 3); if (ok)return;
        if (dx + 1 <4 && map[dx+ 1][dy + 1] == '.'&&map[dx+ 1][dy + 2] == '.'&&map[dx+ 1][dy + 3] == '.') dfs(dx+ 1, dy + 3); if (ok)return;
        
    }
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        memset(vis, 0, sizeof(vis));
        cin >> n >> x;
        int s = 0;
        for (int i = 1; i <= 3; i++)
        {
            cin >> map[i]; if (map[i][0] == 's')s = i;
        }
        ok = 0;

        if (n % 3 != 1)for (int i = 1; i <= 3; i++) { if ((n % 3) == 0)map[i] = map[i] + '.'; if ((n % 3) == 2)map[i] = map[i] + ".."; }
        dfs(s, 0);
        if (ok == 1)cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    //cin >> t;
}