Rescue

最新推荐文章于 2025-07-14 19:25:27 发布

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最新推荐文章于 2025-07-14 19:25:27 发布

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原文链接：http://www.cnblogs.com/get-an-AC-everyday/p/4211222.html

本文提供了一道名为1039:Rescue的算法题的解答思路，采用优先队列结合广度优先搜索的方法，旨在营救被囚禁的朋友Angel，通过遍历监狱地图找到最短路径。

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1039: Rescue

Time Limit: 1 Sec Memory Limit: 32 MB
Submit: 1320 Solved: 306

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

用优先队列+bfs,写得蛮有逻辑性（哈哈）：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<queue>
 5 #define MAX_N 205
 6 using namespace std;
 7 struct Node
 8 {
 9     int x,y,t;
10     bool operator<(const Node &a)const
11     {
12         return t>a.t;
13     }
14 };
15 
16 Node start;
17 
18 char map[MAX_N][MAX_N];
19 
20 int dir[][2]={{0,1},{0,-1},{1,0},{-1,0}};
21 
22 int n,m;
23 
24 int bfs()
25 {
26     priority_queue<Node> que;
27     Node cur,next;
28     int i,j;
29     map[start.x][start.y]='#';
30     que.push(start);
31     while(!que.empty())
32     {
33         cur=que.top();
34         que.pop();
35         for(i=0;i<4;i++)
36         {
37             next.x=cur.x+dir[i][0];
38             next.y=cur.y+dir[i][1];
39             next.t=cur.t+1;
40 
41             if(next.x<0||next.x>=n||next.y<0||next.y>=m)
42                 continue;
43             if(map[next.x][next.y]=='#')
44                 continue;
45             if(map[next.x][next.y]=='r')
46                 return next.t;
47             if(map[next.x][next.y]=='x')
48             {
49                 next.t++;
50                 map[next.x][next.y]='#';
51                 que.push(next);
52             }
53             else if(map[next.x][next.y]=='.')
54             {
55                 map[next.x][next.y]='#';
56                 que.push(next);
57             }
58         }
59     }
60     return -1;
61 }
62 
63 int main()
64 {
65    // freopen("a.txt","r",stdin);
66     int i,ans;
67     char *p;
68     while(scanf("%d%d",&n,&m)!=EOF)
69     {
70         for(i=0;i<n;i++)
71         {
72             scanf("%s",map[i]);
73             if(p=strchr(map[i],'a'))
74             {
75                 start.x=i;
76                 start.y=p-map[i];
77                 start.t=0;
78             }
79         }
80         ans=bfs();
81 
82         printf(ans+1?"%d\n":"Poor ANGEL has to stay in the prison all his life.\n",ans);
83     }
84     return 0;
85 }