HDU - 1072-day9F

问题来源

HDU - 1072

问题描述

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.

输入

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.

输出

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.

例子

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 

output

4
-1
13

解题思路

bfs题 这题的要素是可以回头走，但在几个点上的路可以不用重复，在“4”的点上时间重置，那么就没必要往回走了，还有就是走过的路如果再走那肯定是要在那个要走的路上可以增加时间，用vis数组保存走过那点的最大时间。还有就是剩余时间为1的情况，如果没到终点，也没到“4”点，那就不行了可以去掉那个路径了。

AC的代码

#include<string>
#include<math.h>
#include<queue>
#include<cmath>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include <iostream>
using namespace std;
int xx, yy,ex,ey;
int m, n;
struct lu
{
	int x, y;
	int time=0;
	int stime=6;
};
int dy[] = { 0,1,0,-1 };
int dx[] = { 1,0,-1,0 };
int aa[8][8],vis[8][8];
void dfs()
{
	memset(vis, 0, sizeof(vis));
	queue<lu>q;
	lu wu;
	wu.x = xx;
	wu.y = yy;
	wu.time = 0;
	wu.stime = 6;
	vis[wu.x][wu.y] = 6;
	q.push(wu);
	while (!q.empty())
	{
	
		lu temp;
		temp = q.front();
		q.pop();
		if (temp.x == ex && temp.y == ey)
		{
			cout << temp.time << endl;
			return;
		}
		if (temp.stime == 1)
			continue;
		if (vis[temp.x][temp.y] > temp.stime)
		{
			continue;
		}
		for (int g = 0; g < 4; g++)
		{
			int x1 = temp.x + dx[g];
			int y1 = temp.y + dy[g];
			if (aa[x1][y1] != 0 && x1>=0&&x1<m&&y1>=0&&y1<n)
			{
				if (aa[x1][y1] == 4&&temp.stime>0)
				{
					aa[x1][y1] = 0;
					lu temp1;
					temp1.x = x1;
					temp1.y = y1;
					temp1.stime = 6;
					vis[x1][y1] = temp1.stime;
					temp1.time = temp.time + 1;
					q.push(temp1);
				}
				else
				{
					lu temp1;
					temp1.x = x1;
					temp1.y = y1;
					temp1.stime = temp.stime - 1;
					temp1.time = temp.time + 1;
					vis[x1][y1] = temp1.stime;
					q.push(temp1);
				}
			}
		}

	}
	cout << "-1" << endl;
}
int main()
{
	int t;
	cin >> t;
	for (int q = 0; q < t; q++)
	{
		
		cin >> m >> n;
	
		for (int s =0 ; s <m; s++)
		{
			for (int f = 0; f < n ;f++)
			{
				cin >> aa[s][f];
				if (aa[s][f] == 2)
				{
					xx = s;
					yy = f;
				}
				if (aa[s][f] == 3)
				{
					ex = s;
					ey = f;
				}
			}
		}
		
		dfs();
	}
}