hdu oj 1242 Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14222 Accepted Submission(s): 5164

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

Author

CHEN, Xue

Source

ZOJ Monthly, October 2003

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第一次使用STL的优先队列解决问题，这道题其实就是广搜，只不过把广搜的普通队列换成了priority_queue,单调队列吧。。。其实现在还不大懂 下面奉上AC代码：

#include <iostream>#include <cstdio>#include <queue>using namespace std;const int maxn = 200;int m,n,maze[maxn+10][maxn+10];int dir[][2]={{1,0},{-1,0},{0,1},{0,-1}};typedef struct point{    int x,y,t;    bool operator < (const point &a) const    {        return t>a.t;    }} point;point start;int bfs(){    priority_queue<point>que;    point cur,next;    maze[start.x][start.y]='#';    que.push(start);    while(!que.empty())    {        cur=que.top();        que.pop();        for(int i=0;i<4;i++)        {            next.x=cur.x+dir[i][0];            next.y=cur.y+dir[i][1];            next.t=cur.t+1;            if(next.x<1||next.x>=m+1||next.y<1||next.y>=n+1)                continue;            if(maze[next.x][next.y]=='#')                continue;            if(maze[next.x][next.y]=='r')                return next.t;            if(maze[next.x][next.y]=='.')            {                maze[next.x][next.y]='#';                que.push(next);            }            else if(maze[next.x][next.y]=='x')            {                maze[next.x][next.y]='#';                next.t++;                que.push(next);            }        }    }    return -1;}int main(){    while(~scanf("%d%d",&m,&n))    {    memset(maze,0,sizeof(maze));    for(int i=1;i<=m;i++)        for(int j=1;j<=n;)    {        char c;        scanf("%c",&c);        if(c=='#'||c=='.'||c=='a'||c=='x'||c=='r')        {            maze[i][j++]=c;            if(c=='a')                start.x=i,start.y=j-1,start.t=0;        }    }    int ans=bfs();    printf(ans+1?"%d\n":"Poor ANGEL has to stay in the prison all his life.\n",ans);    }    return 0;}