F - Rescue 优先队列bfs

最新推荐文章于 2019-11-24 11:12:28 发布

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最新推荐文章于 2019-11-24 11:12:28 发布

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原文链接：http://www.cnblogs.com/wzl19981116/p/9498350.html

本文解析了一道经典的POJ算法题目——监狱救援。任务是在一个由墙体、道路和守卫组成的矩阵中找到从朋友位置到被囚禁的Angel位置的最短路径。文章详细介绍了使用广度优先搜索（BFS）算法来解决该问题的方法，并提供了完整的C++代码实现。

来源poj

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

和之前坦克大战一模一样

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define prf(x) printf("%d\n",x) 
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=3e2+10;
char MAP[N][N];
struct node
{
    int x,y,ans;
    friend bool operator <(node a,node b){      return a.ans>b.ans; }
    node(int a=0,int b=0){  x=a;y=b;ans=0;  }
};
int a[4][2]={-1,0,1,0,0,1,0,-1};
priority_queue<node>q;
bool checkt(node a)
{
    return (MAP[a.x][a.y]=='a');
}
int visit[N][N];
int bfs()
{
    node t,tt;
    while(!q.empty())
    {
        t=q.top();
        q.pop();
        visit[t.x][t.y]=1;
        rep(i,0,4)
        {
            tt.ans=t.ans+1;
            tt.x=t.x+a[i][0];
            tt.y=t.y+a[i][1];
            if(!visit[tt.x][tt.y])
            {
                if(checkt(tt))
                return tt.ans;
                if(MAP[tt.x][tt.y]=='.')
                q.push(tt);
                else if(MAP[tt.x][tt.y]=='x')
                {
                    tt.ans++;
                    q.push(tt); 
                }
                visit[tt.x][tt.y]=1;
            }   
        }
    }
    return -1;
}
void display(int n,int m)
{
    mm(visit,0);
    while(!q.empty()) q.pop();
    int x,y;
    rep(i,1,n+1)
    rep(j,1,m+1)
    if(MAP[i][j]=='r')
    { x=i;y=j;}
    q.push(node(x,y)); 
    int ans=bfs();
    if(ans==-1)
    pf("Poor ANGEL has to stay in the prison all his life.\n");
    else
    prf(ans);
}
int main()
{
    int n,m;
    while(~scff(n,m)&&n&&m)
    {
        mm(MAP,'#');
        rep(i,1,n+1)
        {
            sf("%s",MAP[i]+1);
            MAP[i][m+1]='#';
        }
        display(n,m);
    }
    return 0;
}

转载于:https://www.cnblogs.com/wzl19981116/p/9498350.html