hdu 4004 The Frog's Games(二分)

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本文介绍了一只青蛙如何通过有限次数的跳跃成功到达河对岸的问题。文章使用了二分查找算法，针对不同场景进行优化，以找到青蛙所需的最小跳跃能力。

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The Frog's Games

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 8574 Accepted Submission(s): 3992

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

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题意：

一只青蛙要跳到河对岸，在河中有N块石头青蛙可以落脚。青蛙一共可以跳M次，问青蛙最少拥有跳多远的能力能成功跳到河对岸。

思路：

二分答案，时间复杂度为O(N*logL)。

对于每一次判断答案是否成立，需要满足两个条件：

1、每两个石块之间的距离不能长于青蛙跳远距离。

2、青蛙跳的次数最多为m次。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <list>
using namespace std;
const int maxn=500005;
int a[maxn];
int l,n,m;
bool judge(int len)
{
    int cnt=0;
    int sum=0;
    for(int i=1;i<=n+1;i++)
    {
        if(a[i]-a[i-1]>len)
        {
            return 0;
        }
        else
        {
            if(sum+a[i]-a[i-1]>len)
            {
                cnt++;
                sum=a[i]-a[i-1];
            }
            else
                sum=sum+a[i]-a[i-1];
        }
    }
    if(cnt>m-1)
        return 0;
    return 1;
}
int main()
{
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        a[n+1]=l;
        sort(a+1,a+1+n+1);
        int l=1,r=1000000005;
        while(l<r)
        {
            int mid=(l+r)/2;
            if(judge(mid))
            {
                r=mid;
            }
            else
                l=mid+1;
        }
        printf("%d\n",l);
    }
    return 0;
}