杭电1242--Rescue（BFS+优先队列）

最新推荐文章于 2024-09-18 10:20:02 发布

牧野之歌

最新推荐文章于 2024-09-18 10:20:02 发布

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分类专栏： C语言

本文链接：https://blog.youkuaiyun.com/zz_dd_yy/article/details/29211471

这篇博客主要介绍了如何解决杭电1242题目的救援任务问题，利用BFS（广度优先搜索）结合优先级队列来寻找最优路径。代码中定义了节点结构体，通过遍历地图并更新最优时间，最终找到从起点到救援目标的最短时间。如果无法到达目标，则输出提示信息。

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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

# include<stdio.h>
# include<stdlib.h>
# include<string.h>
int num[210][210],n,m,x1,y1,mtime;//num[][]记录走到该点的最优时间
int dir[4][2]={-1,0,1,0,0,-1,0,1};
char map[210][210];
struct node{ 
    
        int x,y,time;
};
struct node queue[500001];
void bfs(struct node start)
{ 
    
        struct node cur,next;
        int head,tail,i;
        head=tail=0;
        queue[tail++]=start;
  