HDU1072 Nightmare(广搜BFS+优先队列)

题目:

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10533    Accepted Submission(s): 5128


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
 

Author
Ignatius.L
 

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代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int go[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
int map[50][50];
//int vis[50][50];
int n,m;
int x1,y1,x2,y2;
struct node
{
    int x,y;
    int step;
    int time;
    friend bool operator < (node a,node b)//重载运算符让步数从小到大排列
    {
        return a.step>b.step;
    }
};
int bfs(int x,int y)
{
    priority_queueq;
    node now,to;
    now.x=x,now.y=y,now.step=0,now.time=6;//初始炸弹时间为6
    map[x][y]=1;//把初始点标记为空地
    q.push(now);
    while(!q.empty())
    {
        now=q.top();
        if(now.x==x2&&now.y==y2&&now.time>0)return now.step;//满足条件时返回步数
        q.pop();
        for(int i=0; i<4; i++)//遍历四个方向
        {
            int xx=now.x+go[i][0];
            int yy=now.y+go[i][1];
            if(xx>=0&&yy>=0&&xx1)//判断越界及时间是否够
            {
                to.x=xx,to.y=yy,to.time=now.time;
                if(map[xx][yy]==4)//遇到时间重置器时重置时间并且标记为墙
                {
                    to.time=6;
                    map[xx][yy]=0;//如果已经走过这个位置了,那么用的时间肯定不是最小的,可以直接标记为0
                }
                else if(map[xx][yy]==1)
                    to.time--;//每当走到一个空地时间减一
                to.step=now.step+1;
                q.push(to);
            }
        }
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        mem(map,0);
        scanf("%d%d",&n,&m);
        for(int i=0; i
题目大衣是:0代表墙,1代表空地,2代表起始位置,3代表终点位置,4代表炸弹重置器位置,炸弹重置器可以重复使用,可以把爆炸时间重置为6秒,在这里我们标记的时候,用过一次炸弹重置器就直接标记为墙,因为如果重复使用炸弹重置器时间肯定不是最短的,而且可能会陷入死循环出不来,走过的路径也不能标记,因为可能需要重复走,那这个问题的关键就变成了标记时间。

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