hdoj 1242 Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19985    Accepted Submission(s):
7110

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

 

　

/*
题意 ：公主被抓（为什么公主老是被魔王抓，烂梗），我要去救公主 （其实我不想救，心累）
a表示公主所在位置，r是我所在位置，#是墙，.是路，X是怪，打怪要两点疲劳值，走路要一点
疲劳值，问救出公主要多少点疲劳值救出公主的话输出疲劳值点数如果救不出输出
 "Poor ANGEL has to stay in the prison all his life."
 题解：同 poj2312（我相信细微的差别你们能改出来）
*/

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 210
using namespace std;
char map[MAX][MAX];
int n,m;
struct node
{
    int x;
    int y;
    int time;

    friend bool operator < (node a,node b)//定义结构体的优先队列
    {
        return a.time>b.time;//花费时间少的先出队
    }    

};
void bfs(int x1,int y1,int x2,int y2)
{
    int j,i,ok=0;

    priority_queue<node>q;//定义结构 体优先队列 

    int move[4][2]={0,1,0,-1,1,0,-1,0};
    node begin,end;
    begin.x=x1;
    begin.y=y1;
    begin.time=0;
    q.push(begin);
    while(!q.empty())
    {

        end=q.top();//优先队列

        q.pop();
        if(end.x==x2&&end.y==y2)
        {
            ok=1;
            break;
        }
        for(i=0;i<4;i++)
        {
            begin.x=end.x+move[i][0];
            begin.y=end.y+move[i][1];
            if(0<=begin.x&&begin.x<n&&0<=begin.y&&begin.y<m&&map[begin.x][begin.y]!='#')
            {
                if(map[begin.x][begin.y]=='x')
                begin.time=end.time+2;
                else
                begin.time=end.time+1;
                map[begin.x][begin.y]='#';
                q.push(begin);

            }
        }
    }
    if(ok)
    printf("%d\n",end.time);//注意这里
    else
    printf("Poor ANGEL has to stay in the prison all his life.\n");
}
int main()
{
    int j,i,s,t,k,x1,x2,y1,y2;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
        scanf("%s",map[i]);
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='a')
                {
                    x2=i;y2=j;
                }
                else if(map[i][j]=='r')
                {
                    x1=i;y1=j;
                }
            }
        }
        bfs(x1,y1,x2,y2);
    }
    return 0;
}