HDU Nightmare 1072 （BFS+有回头路）_nightmare hdu 1072

作者：空白诗007 | 2024-07-28 12:07:59

踩

nightmare hdu 1072

Nightmare

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9610 Accepted Submission(s): 4668

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

Author

Ignatius.L

题解：给你一个图，图中的标志如题所述，人到达4的位置时，炸弹会重置为6，而且， 可以走回头路！！！

所以不可以把走过的路给标志了或者不再走了！！！

AC代码：


#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf 0x6fffff
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
 
int map[10][10];
int vis[10][10];
int n,m,x2,y2;
struct node
{
	int x,y;
	int time,step;
};
void bfs(int x,int y)
{
	int i;
	node st,ed;
	queue<node>q;
	st.x=x;
	st.y=y;
	st.time=6; //炸弹倒时器
	q.push(st);
	while(!q.empty()) 
	{
		st=q.front();
		q.pop();
		if(st.x==x2&&st.y==y2)
		{
			printf("%d\n",st.step);
			return ;
		}
		for(i=0;i<4;i++)
		{
			ed.x=st.x+dir[i][0];
			ed.y=st.y+dir[i][1];
			ed.time=st.time-1; //倒时：减一 
			ed.step=st.step+1; //步数：加一
			if(ed.x>=0 && ed.y>=0 && ed.x<n && ed.y<m && map[ed.x][ed.y]!=0 && ed.time>0 && vis[ed.x][ed.y]<st.time)
		    {
		    	if(map[ed.x][ed.y]==4) //重置炸弹倒时器 
		    	{
		    		ed.time=6; 
				}
				vis[ed.x][ed.y]=ed.time;
				q.push(ed);
				
			}
		}
		
	}
	printf("-1\n");
}
int main()
{
	int t,i,j;
	int x1,y1;
	cin>>t;
	while(t--)
	{
		int i,k;
		cin>>n>>m;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				cin>>map[i][j];
				if(map[i][j]==2)
				{
					x1=i;
					y1=j;
				}
				if(map[i][j]==3)
				{
					x2=i;
					y2=j;
				}
			}
			
		}
		memset(vis,0,sizeof(vis));
			bfs(x1,y1);
	}
	return 0;
}

本文内容由网友自发贡献，转载请注明出处：【wpsshop博客】