hdu 1072 Nightmare(DFS)

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6591    Accepted Submission(s): 3211

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

 

Author

Ignatius.L

 

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题意：就是给你一张最大为8*8的地图，然后2是起点，3是终点，1可以直接通过，0不能通过，到达4时，如果剩余的           时间大于0，可以将炸弹倒计时重置为6，每个点可以重复到达。

分析：刚开始用DFS时没有想清楚，于是各种wa，而且网上搜的题解都是BFS做的，差点就没坚持住直接换BFS了

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int rest[12][12],N,M,ans,flag;int maps[12][12],step[12][12],coun;void dfs(int x,int y,int time,int coun){    //cout<<x<<y<<time<<coun<<endl;    if(maps[x][y]==3&&time>0)    {        if(ans>coun)            ans = coun;        flag =1;        return ;    }    if(time<=0||coun>=ans)//如果当前所剩的时间或者所走的步数超过已经记录的最小到达终点步数，直接pass        return ;    if(maps[x][y] == 4)//碰到4重置为6    {        time =6;    }    if(rest[x][y]>=time&&step[x][y]<=coun)//剪枝：如果曾经到达该节点所剩的时间比现在多并且步数都比现在少，可直接跳过        return ;    rest[x][y] = time;//经过剪枝，这便是记录到达该节点所剩的最大时间和所需的最小步数    step[x][y] = coun;    if(x-1>0&&maps[x-1][y]!=0)    {        dfs(x-1,y,time-1,coun+1);    }    if(x+1<=N&&maps[x+1][y]!=0)    {        dfs(x+1,y,time-1,coun+1);    }    if(y-1>0&&maps[x][y-1]!=0)    {        dfs(x,y-1,time-1,coun+1);    }    if(y+1<=M&&maps[x][y+1]!=0)    {        dfs(x,y+1,time-1,coun+1);    }}int main(){    int sx,sy,T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&M);        flag = 0;ans = (1<<30)-1;        memset(rest,0,sizeof(rest));        for(int i=1;i<=N;i++)        {            for(int j = 1;j<=M;j++)            {                scanf("%d",&maps[i][j]);                if(maps[i][j]==2)                {                    sx = i;                    sy = j;                }                step[i][j] = ans;            }        }        rest[sx][sy] = 6;        dfs(sx,sy,6,0);        if(flag) printf("%d\n",ans);        else printf("-1\n");    }    return 0;}