HDU 1072 Nightmare

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2506    Accepted Submission(s): 1248

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
 
尼条题目大意就系一个人入左去一个迷宫之后，就触发左一个定时6分钟ge炸弹。
距要尽力跑到出口。迷宫有起点（2），空地（1），墙（0），炸弹重置器（4）【定时炸弹重置为6】，出口（3）。
并有如下规则：
1、每分钟，条牟利剩系可以走上下左右四个相邻ge格，并且每步要行一分钟。（唔知系个人行得慢剩系d格太大）
2、如果行到出口ge时候，倒计时又到左0，呢个人就钉柴。
3、如果行到重置器ge时候，倒计时又到左0，呢个人都系钉柴。
4、每个格都可以重复行，（姐系可以无限重置都得，就系用尼条规则剪枝）。
 
我用左广搜+比较按剩余时间剪枝。下面结合代码解释：
 
HDU 1072 0ms 276K 1275B SGetEternal｛（。）（。）｝！
 
#include<iostream>#include<queue>using namespace std;#define INFI (1<<31)-1 struct point //队列类型，step储存步数。{ int x,y,step; void set(int a,int b,int c) { x=a; y=b; step=c; }}h,temp,in;   //h为起始点，temp、in系临时变量
int main(){ int t,n,m,i,j,hx,hy,x,y,mn; char map[10][10]; int tim[10][10]; //储存点(i,j)当前时间 char move[4][2]={0,1,1,0,0,-1,-1,0}; queue<point> que; scanf("%d",&t); while (t--) {  hx=hy=0;  scanf("%d%d",&n,&m);  for (i=0;i<=m+1;i++)   map[0][i]=map[n+1][i]=0;    //设置出界墙,即系系将迷宫外部加墙  for (i=1;i<=n;i++)  {   map[i][0]=map[i][m+1]=0;    //设置出界墙too   for (j=1;j<=m;j++)   {    tim[i][j]=0;    scanf("%d",&map[i][j]);    if (map[i][j]==2) { tim[i][j]=6; hx=i; hy=j; } //起始点剩余时间为6   }  }  mn=INFI;  h.set(hx,hy,0);  que.push(h);  while (!que.empty())  {   temp=que.front();   if (map[temp.x][temp.y]==3 && mn>temp.step)    {    mn=temp.step;    h.step=temp.step;   }   for (i=0;i<4 && tim[temp.x][temp.y]-1;i++)   {    x=temp.x+move[i][0];    y=temp.y+move[i][1];    if (map[x][y]!=0 && tim[x][y]<tim[temp.x][temp.y]-1) //&& 后面就系时间剪枝，当当前点剩余时间-1多余下一点时先行，防回荡，可以自己举例。    {     tim[x][y]=tim[temp.x][temp.y]-1;     if (map[x][y]==4) tim[x][y]=6;     in.set(x,y,temp.step+1);     que.push(in);    }   }   que.pop();  }  if (mn<INFI) printf("%d",h.step);  else printf("-1");  putchar('/n'); } return 0;}
 
精粹就系时间剪枝，吾剪就超时，差好hi远。
9up完毕……