River Hopscotch问题（二分）

又是一个卡了我好久好久好久的二分题
我真的不会二分啊，有没有大佬可以给我讲讲
我就是个憨憨

题目链接River Hopscotch
题目描述：
Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).
To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.
Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).
FJ wants to know exactly how much he can increase the shortest distance before he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

输入要求
Line 1: Three space-separated integers: L, N, and M
Lines 2… N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

输出：
Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

这个题大体意思就是在河中共有n块石头，要求在移走m块石头后，所有石头的最小值最大。

注意：

1.题目输入是多组样例，要使用while(scanf(“%d %d %d”,&l,&n,&m)!=EOF)形式。我就是瞎看不见最后一行的要求嘤嘤嘤

2.输入的石头是距离河岸的长度，要先对数据排序才能进行后续判断。

3.对于二分中间元素的判断：初始low=0，high=河的长度（每一次循环都要更新啊！！！不要像我一样傻乎乎不管low），当两块石头的距离小于中间值时，就移走此时这块，并记录移走石头的数目，当大于中间值时，更新前一块石头的位置，（一定要更新啊！！！）当移走数目大于要求数目时，说明此时间隔取大了，更换high的值。

代码实现

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
long long int a[50010],l,m,n;
int main()
{long long int i,j,low=0,high,mid;while(scanf("%lld %lld %lld",&l,&n,&m)!=EOF){a[0]=0;for(i=1;i<=n;i++){scanf("%d",&a[i]);} high=l;low=0; sort(a,a+n+1);while(low<=high){mid=(low+high)/2;int num=0,last=0;for(i=1;i<=n;i++){if(a[i]-last<mid) num++;else last=a[i];}if(num<=m) low=mid+1;else high=mid-1;}printf("%d\n",high);}return 0;}

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River Hopscotch问题（二分）

注意：

代码实现

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