HDU--1242 -- Rescue [广搜]

 

Rescue

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12409    Accepted Submission(s): 4541

 

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

 

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

 

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

Sample Output

13

 

 

Code:

 

一直WA, WA ,WA

原理是用了优先队列而不是普通的队列

这个原来用过

只是用了friend bool operator< ()

点这里右面链接:  传送门

 

 

#include
#include
#include
using namespace std;

typedef struct point
{
    int x,y;
    int step;
    friend bool operator< (point a,point b)//这里这里这里
    {
        return a.step>b.step;//>号表示出队的是优先级低的,<号表示出队优先级高的
    } 
}point;

char a[211][211];
int vis[211][211];
int n,m,dir[4][2]={1,0,-1,0,0,1,0,-1};
point angel;
int r_x,r_y;

int bfs(point p)
{
    int i;priority_queueq;
    q.push(angel);
    vis[angel.x][angel.y] = 1;
    point head,temp    ;
             
    while(!q.empty())
    {
        head = q.top();    //出队一个并对这个进行搜索 
         q.pop();
        if(head.x==r_x&&head.y==r_y) return head.step;//加上自己    
         for(i=0;i<4;i++)
        {   
           temp.x = head.x + dir[i][0];//temp赋值 
             temp.y = head.y + dir[i][1];   
           if(temp.x<0 || temp.x>=n || temp.y<0 || temp.y>=m) continue;//越界                     
             if(!vis[temp.x][temp.y] && a[temp.x][temp.y] != '#')    
            {    
                if(a[temp.x][temp.y]=='x') temp.step = head.step + 2;//直接把到这一个地方的时间加上 
                   else temp.step = head.step + 1;
                vis[temp.x][temp.y] = 1;
                q.push(temp);        //判断过后再入队 
             }     
        }
    }        
    return -1;
}

int main()
{
    int i,j,t;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i