HDU :1242 Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27990    Accepted Submission(s): 9900

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

解题思路：

简单广搜+优先队列，同类型的还有杭电的 1026题:Ignatius and the Princess

#include
#include
#include
#include
#include

using namespace std;

int N,M,sx,sy;///行列
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
char Map[202][202];
struct pos
{
    int x,y;
    int time;
    friend bool operator < (pos a,pos b)
    {
        return a.time > b.time;
    }
};
void bfs()
{
    int i;
    pos cur,nex;
    cur.x = sx;
    cur.y = sy;
    cur.time = 0;
    priority_queuequ;///优先队列
    Map[sx][sy]='#';
    qu.push(cur);
    while(!qu.empty())
    {
        cur = qu.top();
        qu.pop();
        if(Map[cur.x][cur.y] == 'a')
        {
            printf("%d\n",cur.time);
            return;
        }
        for(i = 0; i < 4; i++)
        {
            nex.x = cur.x + dir[i][0];
            nex.y = cur.y + dir[i][1];
            ///没有出界而且可以行走
            if(nex.x>=0&&nex.x=0&&nex.y