HDUoj 1242 Rescue ( BFS+优先队列

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29289    Accepted Submission(s): 10347

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
    
    
    
    

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

 

Sample Output

    
    
    
    

     
     
     
     13
    
    
    
    

#include 
#include 
#include 
#include 
using namespace std;
#define N 1000
char map[N][N];
bool vis[N][N];
int dir[4][2] = {0,1,0,-1,1,0,-1,0};
int m, n, sx, sy, ex, ey;
struct node {
    int x; int y; int step;
    bool friend operator<(node a,node b) {
        return a.step > b.step;          // 最小堆 
    } 
}st,mid,ne;
bool judge(int x,int y)
{
    if(x>=0 && x<m && y>=0 && yreturn true;
    else
        return false;
 }
int bfs()
{
    priority_queue fq;
    st.x = sx; st.y = sy; st.step = 0;
    fq.push(st);
    vis[sx][sy] = true;
    while(!fq.empty()) {
        mid = fq.top();
        fq.pop();
        if(mid.x==ex && mid.y==ey) return mid.step;
        for(int i = 0;i < 4; i++) {
            ne.x = mid.x + dir[i][0];
            ne.y = mid.y + dir[i][1];
            if(ne.x==ex && ne.y==ey)
                return mid.step+1;
            if(judge(ne.x,ne.y) && map[ne.x][ne.y]!='#' && !vis[ne.x][ne.y]) {
                if(map[ne.x][ne.y] == 'x') 
                    ne.step = mid.step + 2;
                else 
                    ne.step = mid.step + 1;
                vis[ne.x][ne.y] = true;
                fq.push(ne);
            }
        }
    }
return 0;
} 

int main()
{
    while(~scanf("%d%d",&m,&n)) {
        memset(vis,false,sizeof(vis));
        for(int i = 0;i < m; i++) {
            scanf("%s",map[i]);
        }
        for(int i = 0;i < m; i++) {
            for(int j = 0;j < n; j++) {
                if(map[i][j] == 'a') {
                    sx = i, sy = j;
                }
                if(map[i][j] == 'r' ) {
                    ex = i, ey = j;
                }
            }
        }
        int k = bfs();
        if(k)   printf("%d\n",k);
        else    puts("Poor ANGEL has to stay in the prison all his life.");
    }
return 0;
}