POJ3258 二分

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 distanceD_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).

题意：

一条河长度为 L，河的起点(Start)和终点(End)分别有2块石头，S到E的距离就是L。
河中有n块石头，每块石头到S都有唯一的距离
问现在要移除m块石头（S和E除外），每次移除的是与当前最短距离相关联的石头，要求移除m块石头后，使得那时的最短距离尽可能大，输出那个最短距离。
一道比较典型的求最小值最大化的题目
定义函数 c(x) 是求距离x能否留下N-M个石子。然后通过二分找出最大值

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=50005;
int x[maxn];
int L,n,m;
bool C(int d)
{
     int num=n-m;
     int l=0;
     for(int i=0;i<num;i++)
     {
         int r=l+1;
         while(r<=n&&x[r]-x[l]<d)
         {
             r++;
         }
         if(r>n) return false;
         l=r;
     }
     return true;
}
int main()
{
    while(scanf("%d%d%d",&L,&n,&m)!=EOF)
    {
        if(n==m)
        {
            printf("%d
",L);
            return 0;
        }
        x[0]=0;
        x[n+1]=L;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x[i]);
        }
        sort(x,x+n+1);
        int lb=x[1],ub=L;
        while(ub-lb>1)
        {
            int mid=(lb+ub)/2;
            if(C(mid))
            {
                lb=mid;
            }
            else
                ub=mid;
        }
        printf("%d
",lb);
    }
}