Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1235 Accepted Submission(s): 495

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

Recommend Eddy

/*
优先队列是反的
*/
#include
#define N 205
using namespace std;
char mapn[N][N];
int n,m,res;
int visit[N][N];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
struct node
{
    int x,y;
    int val;
    bool operator <(const node &b)const
    {
        return val>b.val;
    }
};
int ok(int x,int y)
{
    if(x<1||x>n||y<1||y>m||visit[x][y]||mapn[x][y]=='#')
        return 1;
    return 0;
}
int bfs(node start,int x,int y)
{
    int i;
    priority_queuefr;
    fr.push(start);
    visit[start.x][start.y]=1;
    //int c=1;
    while(!fr.empty())
    {
        //cout<
        start=fr.top();
        fr.pop();
        //cout<
        if(start.x==x&&start.y==y)
            return start.val;
        //cout<
        for(i=0;i<4;i++)
        {
            node v=start;
            v.x+=dir[i][0];
            v.y+=dir[i][1];
            if(ok(v.x,v.y)) continue;
            v.x=v.x;
            v.y=v.y;
            v.val++;
            if(mapn[v.x][v.y]=='x')
                v.val++;
            fr.push(v);
            visit[v.x][v.y]=1;
        }
    }
    return 0;
}
int main()
{
    // freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(~scanf("%d%d",&n,&m))
    {
        getchar();
        memset(visit,0,sizeof visit);
        node start;
        int ex,ey;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%c",&mapn[i][j]);
                if(mapn[i][j]=='a')
                {
                    ex=i;
                    ey=j;
                }
                if(mapn[i][j]=='r')
                {
                    start.x=i;
                    start.y=j;
                    start.val=0;
                }
            }
            scanf("\n");
        }//输入完成
        //cout<//cout<// for(int i=1;i<=n;i++)
        // {
            // for(int j=1;j<=m;j++)
                // cout<// cout<// }
        int cur=0;
        cur=bfs(start,ex,ey);
        if(!cur)
            puts("Poor ANGEL has to stay in the prison all his life.");
        else
            printf("%d\n",cur);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wuwangchuxin0924/p/5996897.html