HDU--杭电--1072--Nightmare--广搜--这种题目不是简单的标记了，应该是标记过还要能再访问

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5602    Accepted Submission(s): 2784

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

 

#include <iostream>
#include <queue>
using namespace std;
int map[11][11],visit[11][11],n,m,s,x1,y1,x2,y2,xx[4][2]={{-1,0},{1,0},{0,-1},{0,1}};  //妹的看了大爷这么多博客，这些鸟东西就没解释了啊，我们应该心有灵犀了，嘿嘿

当然这次得说明，visit不是记录是否访问了，是记录的这个地址的时间，因为可能走过的路用别的方式到达会时间剩余更多嘛
struct ssss
{
    int x,y,t,step;  //结构体里面有记录的坐标，炸弹在当前剩余的时间，还有的走的步数
}ss;
queue<ssss> q,qq;  //这个也不解释了，说多了怕被别人说我这人活得麻烦
void bfs()
{
    int i,x,y,X,Y,t,step;
    while(!q.empty())
    {
        ss=q.front();q.pop();  //用ss记录队首同时删除队首(这算老招式了，对吧？-  -！)
        X=ss.x;Y=ss.y;t=ss.t-1;step=ss.step+1;  //用X记录ss中的x，Y是ss中的y，终点打记*****t是记录的当前时间还要减一，也就是下一步的时间，step也是一样是下一步的步数
        if(ss.t==0)continue;  //但ss中的t等于零，no好意思，你被炸死了
        else
        {
            if(map[X][Y]==3){s=ss.step;return;}  //听说3是终点嘛，你没经历时间为0被炸死的话这时你就成功脱险了，记录步数并结束
            if(map[X][Y]==4)t=5;  //碰上了4不是可以复位的么？那就复位呗，但是t=5，上面说了，是记录下一个落脚点的时间的
        }
        for(i=0;i<4;i++)
        {
            x=X+xx[i][0];
            y=Y+xx[i][1];
            if(x>=0&&y>=0&&x<n&&y<m&&visit[x][y]<t&&map[x][y]!=0)  //判断是否过界，这样走去(x,y)是否时间更少，是否map是0
            {
                visit[x][y]=t;ss.x=x;ss.y=y;ss.t=t;ss.step=step;q.push(ss);  //visit换时间，把当前坐标(x,y)和时间还有步数打包并入队
            }
        }
    }
}
int main (void)
{
    int i,j,k,l,t;
    cin>>t;
    while(t--&&cin>>n>>m)
    {
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                cin>>map[i][j];visit[i][j]=0;  //初始化visit
                if(map[i][j]==2)x1=i,y1=j;
                if(map[i][j]==3)x2=i,y2=j;
            }
            q=qq;s=-1;visit[x1][y1]=6;  //初始化队列并标记起点时间为6，这样应该没有谁敢跟主角抢这块地了吧？
            ss.x=x1;ss.y=y1;ss.t=6;ss.step=0;q.push(ss);  //打包，入队，不罗嗦
           bfs();
            cout<<s<<endl;
    }
    return 0;
}

有很多人会说错的厉害，具体厉害成神马熊样我不知道，广搜多做几个题后你会发现你的脑子里面就有一个模板，这是自己的，那么碰上生疏一点的题型就得从自己的模板去改，这样最大的好处就是记忆深刻，改起来也犀利得多