hdu1242

제목 링크:
http://acm.hdu.edu.cn/showproblem.php?pid=1242
제목:
Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14614 Accepted Submission(s): 5300
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

Author

CHEN, Xue

Source

ZOJ Monthly, October 2003

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这个题目最开始是没有说清楚。。请看。。

Angel's friends

结果单组就可以过了。。。还有就是这个题目因为碰到守卫会出现时间的不同变化。。所以我运用了剪枝。。。如果步数大于已经确定的步数，则不需要入队。。这个还是挺高效的。。0ms。。。。代码如下：

#include
#include
#include
#define INF 0x3f3f3f3f
const int maxn=200+10;
using namespace std;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
int step[maxn][maxn];
int n,m;
int sum;
char map[maxn][maxn];

struct Map
{
    int x,y;
    int time;
};
queueq;
Map start,end;

int check(int x,int y)
{
    if(x>=1&&x<=n&&y>=1&&y<=m)
      return 1;
    return 0;
}

int bfs()
{
    Map current,next;
    step[start.x][start.y]=0;
    q.push(start);
    while(!q.empty())
    {
        current=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            next.x=current.x+dx[i];
            next.y=current.y+dy[i];
            next.time=current.time+1;
            if(map[next.x][next.y]=='x')
               next.time++;
            if(check(next.x,next.y)&&map[next.x][next.y]!='#'&&next.time

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hdu1242

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