(HDU - 1242)Rescue

最新推荐文章于 2020-02-13 10:35:50 发布

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本文介绍了一道经典的广度优先搜索(BFS)算法题目——Rescue，该题要求在一张包含障碍物、卫兵及道路的N*M网格地图上，计算从天使位置到其朋友位置所需的最短时间。文章详细解析了问题背景、输入输出格式、样例，并给出了使用优先队列实现BFS的C++代码。

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(HDU - 1242)Rescue

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

Total Submission(s): 32353 Accepted Submission(s): 11304

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

题目大意：在一个n行m列的矩阵中，符号a代表天使，r代表天使的朋友，x代表卫兵，.代表路，#代表障碍。一个人上下左右移动一格需要1s，碰到卫兵打倒卫兵又需要1s，即碰到x需要2s。问天使获救最少需要几秒。

思路：典型的bfs。因为天使的朋友可能不止一个，所以要从a出发去找r。又每一个的权值可能不同，所以需要用优先队列。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int maxn=205;
char a[maxn][maxn];
const int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
bool vis[maxn][maxn];
int n,m,sx,sy;

struct node
{
    int x,y,step;
    friend bool operator<(node a,node b)
    {
        return a.step>b.step;
    }
};

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

int bfs()
{
    priority_queue<node> q;
    node start,tmp,next;
    start.x=sx;
    start.y=sy;
    start.step=0;
    vis[sx][sy]=1;
    q.push(start);
    while(!q.empty())
    {
        tmp=q.top();
        q.pop();
        if(a[tmp.x][tmp.y]=='r') return tmp.step;
        for(int i=0;i<4;i++)
        {
            next.x=tmp.x+dir[i][0];
            next.y=tmp.y+dir[i][1];
            if(check(next.x,next.y)&&!vis[next.x][next.y]&&a[next.x][next.y]!='#')
            {
                if(a[next.x][next.y]=='.'||a[next.x][next.y]=='r') next.step=tmp.step+1;
                else next.step=tmp.step+2;
                vis[next.x][next.y]=1;
                q.push(next);
            }
        }
    }
    return -1;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++) scanf("%s",a[i]);
        bool flag=false;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
                if(a[i][j]=='a')
                {
                    sx=i;
                    sy=j;
                    flag=true;
                    break;
                }   
            if(flag) break;
        }   
        memset(vis,0,sizeof(vis));
        int ans=bfs();
        if(ans!=-1) printf("%d\n",ans);
        else printf("Poor ANGEL has to stay in the prison all his life.\n");
    }
    return 0;
}