River Hopscotch（二分最大化最小值）

River Hopscotch

Description

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 D_i from the start (0 < D_i < 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.

Input

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.

Output

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

Sample Input

25 5 2
2
14
11
21
17

Sample Output

4

Hint

Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25)

题解：

河中起点终点各有一个石头，然后再给你N块石头，代表在河中的位置，问去掉M块石头后的最大最小值为多少，二分极大化最小值；二分错点就是关于里面的大于等于之类判断的问题，可以想成当l,r相等时谁最后判断的就好了，l+1跳出来，直接return l-1就可以了；

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define T_T while(T--)
const int MAXN=50010;
int rock[MAXN];
int N,M;
int js(int x){
	int cnt=0;
	int last=rock[0];
	for(int i=1;i<=N+1;i++){
		if(rock[i]-last<x){
			cnt++;
		}
		else last=rock[i];
	}
	return cnt;
}
int erfen(int l,int r){
	int mid;
	while(l<=r){
		mid=(l+r)>>1;
		if(js(mid)>M)r=mid-1;//就把等号去了，后面变成l-1就对了。。。。
		else l=mid+1;
	}
	return l-1;
}
int main(){
	int L;
	while(~scanf("%d%d%d",&L,&N,&M)){
		rock[0]=0;
		for(int i=1;i<=N;i++)SI(rock[i]);
		rock[N+1]=L;
		sort(rock,rock+N+2);
		printf("%d\n",erfen(0,L));
	}
	return 0;
}