The Frog's Games HDU - 4004 二分

最新推荐文章于 2020-03-21 16:17:28 发布

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最新推荐文章于 2020-03-21 16:17:28 发布

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分类专栏：二分

本文链接：https://blog.youkuaiyun.com/mayuqing98/article/details/79808945

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这篇博客探讨了HDU 4004问题，即青蛙如何使用二分搜索策略来确定最小跳跃距离，以确保能在不超过m次跳跃的情况下跨越一系列石头。核心思路是在每个固定跳跃距离下，计算最多能跨越多少次石头，如果次数小于等于m，则该距离是有效的。

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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).

InputThe 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.OutputFor each case, output a integer standing for the frog's ability at least they should have.Sample Input

Sample Output

可以二分青蛙每次的最小跳跃距离。关键在于如何判断这个距离符合题目条件。当青蛙跳跃距离固定时，一次跳尽量多的石头，记录跳的次数，跳跃次数小于等于m则符合条件。

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=500005;
int a[maxn];
int len,m,n;
int mmax;
int judge(int d)
{
    int last=0;
    int cnt=0;
    for(int i=0;i<=n;)
    {
        if(a[i]<=last+d)i++;
        else
        {
            if(last==a[i-1])return 0;                       //如果上次跳到的位置上一块判断的石头相等，说明跳不过去了
            else
            {
                last=a[i-1];                     //可以跳到上一块，即最远满足条件的石头了
                cnt++;
            }
        }
    }
    cnt++;
    if(cnt<=m)return 1;
    else return 0;
}
int erfen()
{
    int l,r,m;
    l=1,r=len;
    int ans=0;
    while(l<=r)
    {
        m=(l+r)/2;

        if(judge(m))
        {
           r=m-1;
           ans=m;

        }
        else l=m+1;
    }
    return ans;
}
int main()
{
   // int len,m,n;
    while(cin>>len>>n>>m)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
       a[n]=len;

        cout<<erfen()<<endl;
    }
    return 0;
}