Problem Description

Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.

Given the layout of the labyrinth and Ignatius' start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.

Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can't get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can't use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.

Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius' start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius' target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.

Output For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.

Sample Input 3 3 3 2 1 1 1 1 0 1 1 3 4 8 2 1 1 0 1 1 1 0 1 0 4 1 1 0 4 1 1 0 0 0 0 0 0 1 1 1 1 4 1 1 1 3 5 8 1 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1
Sample Output 4 -1 13

題目的意思是：給你一個迷宮，0代表牆，1代表路，2代表起始位置，3代表終點，4代表爆破裝置。一開始，你在2的位置，求到3的最少步數。

起初，你只有6秒鐘的時間，時間減到0，你沒到3的位置，代表不能出去，輸出-1.想要增加時間，可以引爆爆破裝置，引爆之後，時間重置為6，可以引爆多個。到引爆裝置的時候，時間必須大於0，不然沒有時間可以引爆。

迷宮最大是8 * 8。用廣度優先搜尋可以簡單的解決問題。

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <queue>  
using namespace std;  
  
typedef struct  
{   
    int Etime;  
    int x, y;  
    int count;  
}Data;  
  
int n, m, sx, sy;  
int map[9][9];  
int direction[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};  
  
int bfs()                       //廣度搜索  
{  
    int i;  
    int tx, ty;  
    queue<Data> Que;  
    while(!Que.empty())  
    {  
        Que.pop();  
    }  
    Data Da, temp;  
    Da.x = sx; Da.y = sy;  
    Da.Etime = 6; Da.count = 0;  
    Que.push(Da);                       //起點入隊  
    while(!Que.empty())  
  {  
        temp = Que.front();  
        Que.pop();  
        if(map[temp.x][temp.y] == 3)  //到了終點  
            return temp.count;  
        if(temp.Etime == 1)           //沒到終點，時間變成1，下一步之後，時間變0，無論怎麼走，都沒時間了，直接跳過  
            continue;             //忽略掉時間為0的，下面的引爆就不用判斷時間  
        for(i = 0; i < 4; i++)        //四個方向搜尋  
        {  
            tx = temp.x + direction[i][0];  
            ty = temp.y + direction[i][1];  
            if(tx < 0 || tx >= n || ty < 0 || ty >= m || map[tx][ty] == 0)  
                continue;  
      
            Da.x = tx; Da.y = ty; Da.Etime = temp.Etime - 1; Da.count = temp.count + 1;   
            if(map[tx][ty] == 4)          //引爆，重置時間  
            {  
                Da.Etime = 6;  
                map[tx][ty] = 0;  
            }  
            Que.push(Da);  
        }  
    }  
    return -1;  
}  
int main()  
{  
//  freopen("data.txt", "r", stdin);  
    int T, i, j;  
    while(scanf("%d", &T) != EOF)  
    {  
        while(T--)  
        {  
            scanf("%d%d", &n, &m);  
            for(i = 0; i < n; i++)  
                for(j = 0; j < m; j++)  
                {  
                    scanf("%d", &map[i][j]);  
                    if(map[i][j] == 2)            //找到起點  
                    {  
                        sx = i;  
                        sy = j;  
                    }  
                }  
            int ans = bfs();  
            cout << ans << endl;  
        }  
    }  
    return 0;  
}