NYOJ 483 Nightmare

Nightmare

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 4

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.

样例输入

2
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

样例输出

4
-1

上传者

ACM_林志强

题目描述：

你身上附有炸弹目前处于2的位置，时间还剩下6分钟，你要在炸弹爆炸之前到达3的位置才能获救，当然在4的位置可以将炸弹的时间重新置6，算是延长时间，0的位置不能走，求出到达3位置最少的步数，否则输出-1。

简单的bfs,代码如下：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

int t;
int n,m;
int a[10][10];
int dx[4] = {-1,1,0,0},dy[4] = {0,0,1,-1};   //四个方向 
struct node
{
int x,y,step,time;
}start;
void bfs()
{
queue<node> q;
q.push(start);
node p1,p2;
while(!q.empty())
{
p1 = q.front();      //p1总是接受队列头的数据，再进行下一步的操作 
q.pop();           //不要忘了 
for(int i = 0;i < 4;i++)
{
p2.step = p1.step + 1;
p2.time = p1.time - 1;
p2.x = p1.x + dx[i];
p2.y = p1.y + dy[i];
if(p2.x>=0 && p2.x<n && p2.y>=0 && p2.y<m && a[p2.x][p2.y]!=0 && p2.time>0)   //p2.time 必须大于0才行 
{
if(a[p2.x][p2.y]==3)
    {
printf("%d\n",p2.step);
      return;
    }
    else if(a[p2.x][p2.y]==4)
    {
      p2.time=6;
      a[p2.x][p2.y]=0;   //考虑到实际情况，访问过4这个位置是不会再次访问的 
    }
    q.push(p2);
}
}
}
printf("-1\n");
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i = 0;i < n;i++)
for(int j = 0;j < m;j++)
{
scanf("%d",&a[i][j]);
if(a[i][j] == 2)      //对初始位置进行初始化 
{
start.x = i;
start.y = j;
start.time = 6;
start.step = 0;
}
}
bfs();
}
return 0;
}