HDUOJ----4004The Frog's Games（二分+简单贪心）

铁蛙三项之跳跃挑战

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原文链接：http://www.cnblogs.com/gongxijun/p/3603607.html

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#java

本文介绍了一项算法挑战——铁蛙三项中的跳跃比赛。通过二分搜索优化方法，确定青蛙完成跳跃所需的最短能力距离。该问题涉及输入处理、排序及条件判断等算法步骤。

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3263 Accepted Submission(s): 1596

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

参考别人的思路做的，挫....

利用二分来选择适合的数。。。然后再验证，是枚举的优化版。好吧，这个思路是别人的，我还是太挫啦

代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<functional>
 5 const int maxn= 500005;
 6 using namespace std;
 7 int stone[maxn],n,m;
 8 bool judge(int dis)  
 9 {
10     int i=1,pre=0,count=0;
11     while(i<=n+1)
12     {
13         count++;
14         if(dis<stone[i]-stone[i-1])  return false ;  //这个石头跳不过去，所以失败
15         while (i<=(n+1)&&dis>=stone[i]-stone[pre]) i++;
16         pre=i-1;
17         if(count>m)   return false ;
18     }
19     return true ;
20 }
21 int main()
22 {
23     int length,i;
24     while(scanf("%d%d%d",&length,&n,&m)!=EOF)
25     {
26          memset(stone,0,sizeof(int)*(n+3));
27           for(i=1;i<=n;i++)
28          scanf("%d",&stone[i]);
29           stone[i]=length;
30        sort(stone,stone+(n+2),less<int>());   //升序安放stone
31        int ans,low=0,high=length;
32         while(low<=high)
33         {
34              ans=(low+high)/2;   //跳这么远的时候能否满足要求
35             if(ans*m>=length&&judge(ans))  high=ans-1;
36             else low=ans+1;
37         }
38         printf("%d\n",low);
39     }
40     return 0;
41 }