HDU 1072 Nightmare (whether he can leave before the bomb explosion with a time bomb-BFS + DP)

HDU 1072 Nightmare (whether he can leave before the bomb explosion with a time bomb-BFS + DP)

Nightmare Time Limit:1000 MS Memory Limit:32768KB 64bit IO Format:% I64d & % I64u

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 shocould 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 cocould 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 cocould reset the exploding time to 6 minutes.

Given the layout of the labyrinth and Ignatius 'start position, please tell Ignatius whether he cocould get out of the labyrinth, if he cocould, 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 cocould only get to one of the nearest area, and he shocould not walk out of the border, of course he cocould 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 frequently times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as frequently 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 wocould 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 shocould 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 shocould output the minimum time he needs, else you shoshould 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 

Question:

A person is in the Maze and carries a time bomb. After six seconds, the bomb will explode. There are five kinds of squares in the Maze. 2 represents the starting point, and 3 represents the exit, 1 indicates the road that can be taken, 0 indicates the wall, and 4 indicates that the current time bomb can be reset to 6 at this point. PS: the exit condition is that the remaining time of the bomb is greater than 0, and the condition for resetting the bomb time is that the explosion time of the Current bomb is greater than 0.

Solution:

BFS + DP, when the next step can take more time than the actual step, update the explosion time ..

Code:

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std;int dirI[4]={1,0,-1,0},is,js,ie,je,n,m;int dirJ[4]={0,1,0,-1},timen[10][10];struct node{    int i,j,t,v;    node(int i0=0,int j0=0,int t0=0,int v0=0){        i=i0,j=j0,t=t0,v=v0;    }}a[10][10];void bfs(){    queue 
     
       path;    timen[is][js]=0;    a[is][js].t=6;    bool ans=false;    path.push(a[is][js]);    while(!path.empty()&&!ans){        node s=path.front();        path.pop();        for(int i=0;i<4;i++){            int di=s.i+dirI[i],dj=s.j+dirJ[i];            if(a[di][dj].v==0||di<0||dj<0||di>=n||dj>=m) continue;            int temp=timen[s.i][s.j]+1,st=s.t-1 ;            if(st>a[di][dj].t||timen[di][dj]==-1){                if(a[di][dj].v==4&&st>=1) {st=6;a[di][dj].t=st;timen[di][dj]=temp;path.push(a[di][dj]);}                if(a[di][dj].v==1) {a[di][dj].t=st;timen[di][dj]=temp;path.push(a[di][dj]);}                if(a[di][dj].v==3&&st>=1){ans=true;timen[di][dj]=temp;break;}            }        }    }    printf("%d\n",timen[ie][je]);}int main(){    int T;    scanf("%d",&T);    while(T--){        memset(timen,-1,sizeof(timen));        scanf("%d%d",&n,&m);        for(int i=0;i