hdu 4004 The Frog's Games 二分

最新推荐文章于 2025-09-06 09:33:29 发布

weixin_34234721

最新推荐文章于 2025-09-06 09:33:29 发布

阅读量89

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/tom987690183/p/3562294.html

本文介绍了一道经典的算法竞赛题目——铁蛙三项中的跳跃挑战。任务是在限定跳跃次数内，计算青蛙至少需要具备的跳跃能力才能成功越过河流。文章提供了完整的C++代码实现，并采用二分查找的方法来优化解决方案。

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3050 Accepted Submission(s): 1505

Problem Description

The annual Games in frogs' kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog's longest jump distance).

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.

Output

For each case, output a integer standing for the frog's ability at least they should have.

Sample Input

   6 1 2 
  

2

25

3 3

11

2

18

Sample Output

4

11

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 1 /*
 2 题意：L, n , m 分别代表河的长度,n个石头,最多能跳m次。
 3           求在满足跳的步数的前提下，到达河对岸的最小步伐。
 4           两个二分比较容易理解。
 5           第一个二分，[ 0, L ];
 6           第二个二分用来在当前步伐下能不能在规定步数内
 7           走到对岸时的，寻找下一个位置。(下限值).
 8           二分写得很挫。
 9 
10 */
11 
12 #include<iostream>
13 #include<stdio.h>
14 #include<cstring>
15 #include<cstdlib>
16 #include<algorithm>
17 using namespace std;
18 
19 int L,n,m;
20 int a[500003];
21 
22 int EF(int l,int r,int num)
23 {
24     int mid;
25     while(l<r)
26     {
27         mid=(l+r+1)/2;
28         if( a[mid]>num)
29             r=mid-1;
30         else if( a[mid]<num)
31             l=mid;
32         else if( a[mid]==num)
33             return mid;
34     }
35     return r;
36 }
37 int query(int length,int l,int r)
38 {
39     int num=0,cur=1,x=0;
40     while( ++num<=m&&a[cur]<L)
41     {
42         cur=EF(cur,n+1,a[x]+length);
43         x=cur;
44     }
45     return a[cur];
46 }
47 int main()
48 {
49     int i,l,r,mid;
50     while(scanf("%d%d%d",&L,&n,&m)>0)
51     {
52         for(i=1;i<=n;i++)
53             scanf("%d",&a[i]);
54         a[n+1]=L;a[0]=0;
55         l=0;r=L;
56         sort(a+1,a+1+n);
57         while(l<r)
58         {
59             mid=(l+r)/2;
60             if( query(mid,1,n+1)>=L )
61                 r=mid;
62             else l=mid+1;
63         }
64         printf("%d\n",r);
65     }
66     return 0;
67 }