hdu1072 Nightmare(bfs)

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9662    Accepted Submission(s): 4696

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

33 32 1 11 1 01 1 34 82 1 1 0 1 1 1 01 0 4 1 1 0 4 11 0 0 0 0 0 0 11 1 1 4 1 1 1 35 81 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

4-113

 

题意：给你一个图，1代表路，0代表墙，2代表起始点，3代表终点，4可以重置炸弹时间，炸弹爆炸时间为6min，每走一步需要1min，如果时间变成0就算走到了3或者4也算死亡。问你能否走出，能输出步数，不能输出-1。

思路： 题目中有一句：如果你需要，可以重复到4这个点补充炸弹时间。其实并没有什么用，你经过了某一个4，根据广搜如果走不出的话无论你怎么回头再走这同一个4也是走不出去的，所以直接将如果经过一个4那么就把这个4设为0,就可以了。         还有一个最重要的问题，根据普通的广搜，我们有vis数组，走过的点都会标记成不能在走，但这个题如果经过了4点重置时间后，是很有可能需要走重复点的，为了解决这个就要在vis上作出改变了，我们只需要用vis记录每次经过一点时候boom所剩余的时间，如果两次经过某一点剩余时间不同，说明时间得到了补充，那么就可以重复走。

Code：

/*******************************************FileName:bfsAuthor:苏相学Date:2016年6月26日15:44:37Description:功能Others:********************************************/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <algorithm>#include <queue>using namespace std;const int maxn = 105;struct node{    int x,y,time;}nex,now;int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};int map[120][120];int vis[120][120];int output[120][120];int m,n;void bfs(int x,int y){    queue<node> s;    memset(vis,0,sizeof(vis));    memset(output,0,sizeof(output));    vis[x][y]=6;    now.x=x;    now.y=y;    now.time=6;    s.push(now);    while(!s.empty())    {        now=s.front();        s.pop();        for(int i=0;i<4;i++){            nex.x=now.x+fx[i];            nex.y=now.y+fy[i];            nex.time=now.time;            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&vis[nex.x][nex.y]!=nex.time&&map[nex.x][nex.y]!=0)            {                    if(map[nex.x][nex.y]==4)                    {                        nex.time=now.time-1;                        if(nex.time==0) continue;                        output[nex.x][nex.y]=output[now.x][now.y]+1;                        vis[nex.x][nex.y]=nex.time;                        nex.time=6;                        s.push(nex);                        map[nex.x][nex.y]=0;                    }                    if(map[nex.x][nex.y]==1)                    {                        output[nex.x][nex.y]=output[now.x][now.y]+1;                        nex.time=now.time-1;                        vis[nex.x][nex.y]=nex.time;                        if(nex.time==0) continue;                        s.push(nex);                    }                    if(map[nex.x][nex.y]==3)                    {                        nex.time=now.time-1;                        if(nex.time==0) continue;                        printf("%d\n",output[now.x][now.y]+1);                        return ;                    }            }        }    }    printf("-1\n");    return ;}int main(){    int t,i,j;    scanf("%d",&t);    while(t--)    {        int x,y;        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)                {                    x=i;                    y=j;                }            }         bfs(x,y);    }    return 0;}

这个题算是需要一点技巧的bfs了，参考了大牛的博客才慢慢理解，也算是练习了一下吧。