hdu 1548 A strange lift (dijkstra)

最新推荐文章于 2021-08-10 15:14:53 发布

转载最新推荐文章于 2021-08-10 15:14:53 发布 · 90 阅读

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原文链接：http://www.cnblogs.com/GetcharZp/p/9438542.html

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本文探讨了一个具有特殊楼层跳跃功能的电梯，通过使用Dijkstra算法解决从任意起点到达终点所需的最少按钮按压次数的问题。输入包含多个测试用例，每个用例描述了电梯在不同楼层的跳跃能力，目标是找到从A楼层到B楼层的最短路径。

A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32529 Accepted Submission(s): 11664

Problem Description
There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?

Input
The input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.

Output
For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".

Sample Input
5 1 5
3 3 1 2 5
0

Sample Output
3

C/C++:

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <string>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <climits>
 8 #include <algorithm>
 9 #define INF 0xffffff
10 using namespace std;
11 
12 int n, my_begin, my_end, my_map[210][210];
13 
14 int my_dijkstra()
15 {
16     int my_book[210] = {0}, my_dis[210];
17     my_book[my_begin] = 1;
18     for (int i = 1; i <= n; ++ i)
19         my_dis[i] = my_map[my_begin][i];
20     while (1)
21     {
22         int my_pos = -1, my_min = INF;
23         for (int i = 1; i <= n; ++ i)
24         {
25             if (!my_book[i] && my_dis[i] < my_min)
26             {
27                 my_min = my_dis[i];
28                 my_pos = i;
29             }
30         }
31         if (my_pos == -1) break;
32         my_book[my_pos] = 1;
33         for (int i = 1; i <= n; ++ i)
34             if (!my_book[i])
35                 my_dis[i] = min(my_dis[i], my_dis[my_pos] + my_map[my_pos][i]);
36     }
37     if (my_dis[my_end] == INF) return -1;
38     return my_dis[my_end];
39 }
40 
41 int main()
42 {
43     while (~scanf("%d", &n), n)
44     {
45         for (int i = 1; i <= n; ++ i)
46             for (int j = 1; j <= n; ++ j)
47                 my_map[i][j] = i == j ? 0 : INF;
48 
49         scanf("%d%d", &my_begin, &my_end);
50         for (int i = 1; i <= n; ++ i)
51         {
52             int temp;
53             scanf("%d", &temp);
54             if (i - temp >= 1) my_map[i][i-temp] = 1;
55             if (i + temp <= n) my_map[i][i+temp] = 1;
56         }
57         printf("%d\n", my_dijkstra());
58     }
59     return 0;
60 }

转载于:https://www.cnblogs.com/GetcharZp/p/9438542.html