Rescue（BFS+优先队列）

BFS+优先队列救援任务

最新推荐文章于 2023-07-26 19:18:14 发布

Chien d'amis

最新推荐文章于 2023-07-26 19:18:14 发布

阅读量170

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM 文章标签： C++

本文链接：https://blog.youkuaiyun.com/u012310622/article/details/97765656

ACM 专栏收录该内容

70 篇文章

订阅专栏

本博客介绍了一个利用BFS（广度优先搜索）和优先队列解决的救援问题，目标是在一个包含墙壁、道路和守卫的监狱地图上找到从朋友位置到被囚禁的天使位置的最短路径。文章详细解释了算法实现，包括如何处理敌人和多组输入的情况。

Rescue（BFS+优先队列）

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

本题需要使用BFS+优先队列。BFS查找路径，优先队列保证所走路程绝对是最短的。
ps.本题要求多组输入，英语不好，当时没注意这个问题，一直wrong answer。

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

char map[201][201]; //储存地图
int visit[201][201];  //标记走过的点
int n, m;  //读取行列
int x1, y1, x2, y2;  //读取初始位置和终点位置
int dir[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };  //遍历右左上下四个方向

struct node
{
	int x, y, k;
	friend bool operator < (node a,node b)  //重载运算符，保证优先队列中，总路程短的在队头
	{
        return a.k >b.k;
    }
};

int check(int x, int y)  //检查走过的点是否满足题意，即：不会走出地图，也不会有障碍物
{
	if (x >= 0 && x < n && y >= 0 && y < m&&map[x][y]!='#')
		return 1;
	return 0;
}

int bfs()
{
	node now, next;
	priority_queue <node> que;  //优先队列
	now.x = x1;
	now.y = y1;
	now.k = 0;
	visit[x1][y1] = 1;
	que.push(now);
	while (que.size())
	{
		now = que.top(); que.pop();
		if (now.x == x2 && now.y == y2)  //若走到终点则弹出优先队列中，总路程最短的那个路程。
		{
			return now.k;
		}
		for (int i = 0; i < 4; i++)
		{
			int x = now.x + dir[i][0];
			int y = now.y + dir[i][1];
			if (check(x, y))
			{
				if (!visit[x][y])
				{
					next.x = x;
                    next.y = y;
                    if(map[x][y]=='x')   //核心判断，若是敌人，杀死后，路程+2
                        next.k=now.k+2;
                    else                        //若是普通路程，路程+1
                        next.k=now.k+1;
					visit[x][y] = 1;
					que.push(next);
                }
			}
		}
	}
	return 0;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)  //这里是坑，题目要求多组输入，一开始没翻译出来。
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                cin >> map[i][j];
        memset(visit, 0, sizeof(visit));
        for (int i = 0; i < n; i++)       //找出初始位置与终点位置
            for (int j = 0; j < m; j++)
            {
                if(map[i][j]=='r')
                {
                    x1=i;
                    y1=j;
                }
                if(map[i][j]=='a')
                {
                    x2=i;
                    y2=j;
                }
            }
        int step=bfs();
        if (step)
            cout << step << endl;
        else
            cout << "Poor ANGEL has to stay in the prison all his life." << endl;
    }
    return 0;
}