Rescue(Hdu 1242)

Rescue
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
 

AC代码:

 

     

#include
#include
#include
#include
#include
#include
#include
#include
#include

typedef long long ll;
const int inf = 0x3f3f3f3f;

using namespace std;

#define maxsize 200

char e[205][205];
int book[205][205];
int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};
int n,m,ans,startx,starty;

void dfs(int a,int b,int cur){        //有多个r 应该从 a出发寻找最近的r

    int ta,tb;

    if(cur >= ans) return ;

    if(e[a][b] == 'r')
    {
        
        ans = cur;

        return ;
    }

    for(int i = 0;i < 4;i++)
    {

        ta = a + dir[i][0];
        tb = b + dir[i][1];
        if(ta >= 0 && tb >= 0 && ta < n && tb < m && !book[ta][tb] && e[ta][tb] != '#')
        {
            book[ta][tb] = 1;
            if(e[ta][tb] == 'x') dfs(ta,tb,cur + 2);
            else dfs(ta,tb,cur + 1);
            book[ta][tb] = 0;
        }

    }

    return ;

}

int main(){

    while(cin >> n >> m)
    {

        getchar();
        int t = 0;
        ans = inf;
        memset(book,0,sizeof(book));

        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < m;j++)
            {
                scanf("%c",&e[i][j]);
                if(e[i][j] == 'a')
                {
                    startx = i;
                    starty = j;
                }
            }
            getchar();    
        }

        book[startx][starty] = 1;
        dfs(startx,starty,0);
        
        if(ans == inf) printf("Poor ANGEL has to stay in the prison all his life.\n");
        else printf("%d\n",ans);
    }

    return 0;

}