(中等) CF 585B Phillip and Trains，BFS。

　　The mobile application store has a new game called "Subway Roller".

The protagonist of the game Philip is located in one end of the tunnel and wants to get out of the other one. The tunnel is a rectangular field consisting of three rows and n columns. At the beginning of the game the hero is in some cell of the leftmost column. Some number of trains rides towards the hero. Each train consists of two or more neighbouring cells in some row of the field.

All trains are moving from right to left at a speed of two cells per second, and the hero runs from left to right at the speed of one cell per second. For simplicity, the game is implemented so that the hero and the trains move in turns. First, the hero moves one cell to the right, then one square up or down, or stays idle. Then all the trains move twice simultaneously one cell to the left. Thus, in one move, Philip definitely makes a move to the right and can move up or down. If at any point, Philip is in the same cell with a train, he loses. If the train reaches the left column, it continues to move as before, leaving the tunnel.

Your task is to answer the question whether there is a sequence of movements of Philip, such that he would be able to get to the rightmost column.

 

　　题意就是问能不能从左到右，假设车不动，然后人向右走，一次一步，分析可知只有当人在的行数是mod 3=2时才能向上下，否则只能向右。

　　所以就可以直接BFS。

　　但是还要注意一个问题就是如果上一步是上下那么这一步就不能是，因为这个被hack了，好不爽。。。

 

代码如下：

// ━━━━━━神兽出没━━━━━━
//      ┏┓       ┏┓
//     ┏┛┻━━━━━━━┛┻┓
//     ┃           ┃
//     ┃     ━     ┃
//     ████━████   ┃
//     ┃           ┃
//     ┃    ┻      ┃
//     ┃           ┃
//     ┗━┓       ┏━┛
//       ┃       ┃
//       ┃       ┃
//       ┃       ┗━━━┓
//       ┃           ┣┓
//       ┃           ┏┛
//       ┗┓┓┏━━━━━┳┓┏┛
//        ┃┫┫     ┃┫┫
//        ┗┻┛     ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━

// Author        : WhyWhy
// Created Time  : 2015年10月12日 星期一 18时20分55秒
// File Name     : D.cpp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

const int MaxN=110;

int N,K;

char map1[5][MaxN];
int vis[5][MaxN];

int que[1000000];
int first,last;

void bfs()
{
    first=last=0;
    int p;
    int x,y;
    int t;
    int u;
    for(int i=1;i<=3;++i) if(map1[i][1]=='s') p=i;

    que[last++]=p*1000+1;
    vis[p][1]=1;

    while(last-first)
    {
        u=que[first++];
        x=u/1000;
        y=u%1000;
        t=vis[x][y];

        if(y<N && vis[x][y+1]!=1 && map1[x][y+1]=='.')
        {
            vis[x][y+1]=1;
            que[last++]=x*1000+y+1;
        }

        if(y%3==2 && x>1 && t==1 && vis[x-1][y]==0 && map1[x-1][y]=='.')
        {
            vis[x-1][y]=2;
            que[last++]=(x-1)*1000+y;
        }

        if(y%3==2 && x<3 && t==1 && vis[x+1][y]==0 && map1[x+1][y]=='.')
        {
            vis[x+1][y]=2;
            que[last++]=(x+1)*1000+y;
        }
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int T;
    scanf("%d",&T);

    while(T--)
    {
        scanf("%d %d",&N,&K);
        for(int i=1;i<=3;++i)
            scanf("%s",map1[i]+1);
        memset(vis,0,sizeof(vis));

        bfs();

        if(vis[1][N] || vis[2][N] || vis[3][N])
            puts("YES");
        else puts("NO");
    }
    
    return 0;
}