【33.33%】【codeforces 586D】Phillip and Trains

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Input

Each test contains from one to ten sets of the input data. The first line of the test contains a single integer t (1 ≤ t ≤ 10 for pretests and tests or t = 1 for hacks; see the Notes section for details) — the number of sets.

Then follows the description of t sets of the input data.

The first line of the description of each set contains two integers n, k (2 ≤ n ≤ 100, 1 ≤ k ≤ 26) — the number of columns on the field and the number of trains. Each of the following three lines contains the sequence of n character, representing the row of the field where the game is on. Philip’s initial position is marked as ‘s’, he is in the leftmost column. Each of the k trains is marked by some sequence of identical uppercase letters of the English alphabet, located in one line. Distinct trains are represented by distinct letters. Character ‘.’ represents an empty cell, that is, the cell that doesn’t contain either Philip or the trains.

Output

For each set of the input data print on a single line word YES, if it is possible to win the game and word NO otherwise.

Examples

input

2

16 4

…AAAAA……..

s.BBB……CCCCC

……..DDDDD…

16 4

…AAAAA……..

s.BBB….CCCCC..

…….DDDDD….

output

YES

NO

input

2

10 4

s.ZZ……

…..AAABB

.YYYYYY…

10 4

s.ZZ……

….AAAABB

.YYYYYY…

output

YES

NO

Note

In the first set of the input of the first sample Philip must first go forward and go down to the third row of the field, then go only forward, then go forward and climb to the second row, go forward again and go up to the first row. After that way no train blocks Philip’s path, so he can go straight to the end of the tunnel.

Note that in this problem the challenges are restricted to tests that contain only one testset.

【题目链接】:http://codeforces.com/contest/586/problem/D

【题解】



bfs题；

首先从每辆车开始处理出每个点在某时刻会不会被车占据;

即can[x][y][t];

其中x最大为3,y最大为100,t最大也为100；

空间没问题;

然后模拟一下每辆车从右到左行驶的情况就可以了(不要忘记一开始的0时刻这个位置也被占据了);

然后从起点开始bfs;

要往右走的时候先看看右边那个位置有没有被车占据;

如果被车占据了；则什么都不能做;表示这个状态是不可行的(当然只是说这个状态不可行，可能还有其他状态在扩展，所以不能直接输出无解);

如果没有被车占据;那么就尝试往右走;时间+1，然后看看在这个时间下会不会被车撞;如果不会被车撞则把这个状态加入到队尾，并标记这个状态被占据(不加这个剪枝会各种T和WA);

如果右边那个位置没有被占据；则可以先往右再往上；到了上面再把时间+1;看看在这个位置会不会被撞.不行的话…..

右下同理;



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long

using namespace std;

const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
struct abc
{
    int x,y,t;
};

int t,n;
bool can[4][110][110];
queue <abc> dl;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

bool bfs(int x,int y)
{
    can[x][y][0] = true;
    while (!dl.empty()) dl.pop();
    abc temp;
    temp.x = x,temp.y = y,temp.t = 0;
    dl.push(temp);
    while (!dl.empty())
    {
        int q1 = dl.front().x,q2 = dl.front().y,t = dl.front().t+1;
        dl.pop();
        if (q2 == n)
            return true;
        q2++;
        abc temp1;
        if (!can[q1][q2][t-1] && !can[q1][q2][t])
        {
            temp1.x = q1,temp1.y = q2,temp1.t = t;
            can[q1][q2][t] = true;
            dl.push(temp1);
        }
        if (!can[q1][q2][t-1] && q1-1>=1 && !can[q1-1][q2][t-1] && !can[q1-1][q2][t])
        {
            temp1.x = q1-1,temp1.y = q2,temp1.t = t;
            can[q1-1][q2][t] = true;
            dl.push(temp1);
        }
        if (!can[q1][q2][t-1] && q1+1<=3 && !can[q1+1][q2][t-1] && !can[q1+1][q2][t])
        {
            temp1.x = q1+1,temp1.y = q2,temp1.t = t;
            can[q1+1][q2][t] = true;
            dl.push(temp1);
        }
    }
    return false;
}

int main()
{
    rei(t);
    while (t--)
    {
        memset(can,false,sizeof(can));
        int sx,sy,m;
        rei(n);rei(m);
        string s1;
        for (int i = 1;i <= 3;i++)
        {
            cin>>s1;
            for (int j = 1;j <= n;j++)
                {
                    if (s1[j-1]=='s')
                        sx = i,sy = j;
                    else
                        if (isalpha(s1[j-1]))
                        {
                            int l = j-1,r = j-1;
                            while (r+1<=n-1 && isalpha(s1[r+1]))
                                r++;
                            l++;r++;
                            int tempr = r;
                            int t = 0;
                            for (int k = l;k <= r;k++)
                                can[i][k][0] = true;
                            while (r>1)
                            {
                                t++;
                                l--;if (l <1) l = 1;
                                r--;
                                l--;if (l <1) l = 1;
                                r--;if (r <1) break;
                                for (int k = l;k <= r;k++)
                                    can[i][k][t] = true;
                                can[i][l][t] = true;
                            }
                            j = tempr;
                        }
                }
        }
        if (bfs(sx,sy))
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}