【二分搜索】最大化最小值:POJ2456

题目：POJ2456

Aggressive cows
Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36441		Accepted: 16640
Description

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?
Input

* Line 1: Two space-separated integers: N and C

* Lines 2..N+1: Line i+1 contains an integer stall location, xi
Output

* Line 1: One integer: the largest minimum distance
Sample Input

5 3
1
2
8
4
9
Sample Output

3
Hint

OUTPUT DETAILS:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.

Huge input data,scanf is recommended.
Source

USACO 2005 February Gold

这种最大化最小值或者最小化最大值的问题，通常可以使用二分搜索解决。

我们定义条件：可以安排牛的位置使得最近两头牛的距离不小于d

然后就通过二分搜索的方式，求解出满足条件的最小的d，就是答案了。

AC代码

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

#define MAXN 100005

int n, c;
ll x[MAXN];

bool check(ll d)
{
    int tmpc = c - 1;
    int i = 1;
    ll last_pos = x[0];
    while (tmpc && i < n)
    {
        if (last_pos + d <= x[i])
        {
            //当前牛舍放牛
            --tmpc;
            last_pos = x[i];
        }
        ++i;
    }
    if (tmpc)
        return false;
    else
        return true;
}

void solve()
{
    ll l = 0, r = 1e9 + 7;
    ll mid;
    while (l+1< r)
    {
        mid = ((l + r) >> 1);
        if (check(mid))
        {
            l = mid;
        }
        else
            r = mid;
            
    }
    printf("%lld\n", l);
}

int main()
{
    scanf("%d%d", &n, &c);
    for (int i = 0; i < n; ++i)
    {
        scanf("%lld", &x[i]);
    }

    sort(x, x + n);
    solve();
}

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