POJ 3258 River Hopscotch (二分,最大化最小值)

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 Mrocks (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 ofM 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).

有N+2块石头，给出最后一块到第一块的距离L，输入中间N块石头到第一块石头的距离，要求移除(除了第一块和最后一块之外的)M块石头，使相邻两块石头的距离中的最小值尽可能得大。
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
int rock[50010],temp[50010];
int main()
{
    int L,N,M;
    cin>>L>>N>>M;
    rock[0]=0;
    rock[N+1]=L;
    for(int i=1;i<=N;i++)
        scanf("%d",&rock[i]);
    N=N+2;
    sort(rock,rock+N);
    for(int i=N-1;i>=1;i--)//N是石块总个数(包含起点,终点),N-1才是最后一块的下标
        rock[i]=rock[i]-rock[i-1];//计算第i块石头与第i-1块石头之间的距离
    int low=0,high=L,mid;
    while(low<high)
    {
        for(int i=1;i<N;i++)//临时存储变量,每一轮都要重新赋值
            temp[i]=rock[i];
        mid=(low+high)/2;
        int cnt=0;//统计在该mid值下可以移除石头的个数
        for(int i=1;i<N;i++)
        {
            if(temp[i]<=mid)
            {
                cnt++;//移除石块的个数加1
                temp[i+1]=temp[i+1]+temp[i];//如果我们移除了第i块石头,那么第i+1块石头的前一块就变成了第i-1块
            }
        }
        if(cnt<=M)//说明mid太小了
            low=mid+1;
        else
            high=mid;
    }
    cout<<high<<endl;
    return 0;
}