HDOJ题目3309 Roll The Cube（BFS）

Roll The Cube

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 502    Accepted Submission(s): 181

Problem Description

This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up. 
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.

 

Input

First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).

 

Output

The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.

 

Sample Input

46 3****B**B**H**H****4 4*****BB**HH*****4 4*****BH**HB*****5 6*******.BB***.H*H**..*.*******

 

Sample Output

312Sorry , sir , my poor program fails to get an answer.

 

Author

MadFroG

 

Source

HDOJ Monthly Contest – 2010.02.06

 

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ac代码

#include<stdio.h>#include<string.h>#include<iostream>#include<queue>using namespace std;int n,m,vis[25][25][25][25],sx[2],sy[2];char map[25][25];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};struct s{int x[2],y[2],step,b[2],h[2];//friend bool operator <(s a,s b)//{//return a.step>b.step;//}}a,temp;int bfs(){memset(vis,0,sizeof(vis));a.x[0]=sx[0],a.x[1]=sx[1];a.y[0]=sy[0],a.y[1]=sy[1];a.b[0]=a.b[1]=a.h[0]=a.h[1]=0;vis[sx[0]][sy[0]][sx[1]][sy[1]]=1;a.step=0;//priority_queue<struct s>q;queue<struct s>q;q.push(a);while(!q.empty()){int i,j;//a=q.top();a=q.front();q.pop();for(i=0;i<4;i++){temp=a;for(j=0;j<2;j++){if(temp.b[j])continue;temp.x[j]=a.x[j]+dx[i];temp.y[j]=a.y[j]+dy[i];if(map[temp.x[j]][temp.y[j]]=='*'){temp.x[j]=a.x[j];temp.y[j]=a.y[j];}}if(vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]])continue;if(temp.x[0]==temp.x[1]&&temp.y[0]==temp.y[1]&&temp.b[0]+temp.b[1]==0)continue;vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]]=1;temp.step=a.step+1;int flag=1;for(j=0;j<2;j++){int now=map[temp.x[j]][temp.y[j]];if(now<2&&!temp.h[now]){temp.h[now]=1;temp.b[j]=1;}if(!temp.b[j])flag=0;}if(flag)return temp.step;q.push(temp);}}return -1;}int main(){int t;scanf("%d",&t);while(t--){int i,j;scanf("%d%d",&n,&m);int cnt=0,cot=0;for(i=0;i<n;i++){scanf("%s",map[i]);for(j=0;j<m;j++){if(map[i][j]=='H'){map[i][j]=cnt++;}if(map[i][j]=='B'){sx[cot]=i;sy[cot]=j;cot++;}}}int ans=bfs();if(ans==-1){printf("Sorry , sir , my poor program fails to get an answer.\n");}elseprintf("%d\n",ans);}}