POJ 2456 Aggressive cows

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

通過二分實現最大化最小值。
思路通過二分的方式來不斷壓縮最小值的可取區間，然後通過貪心的思想去檢查。

#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint N, M;int x[100005];bool C(int d){    int last = 0;    for (int i = 1; i < M; i++)    {        int crt = last + 1;        while (crt < N && (x[crt] - x[last]) < d)        {            crt++;        }        if (crt == N)            return false;        last = crt;    }    return true;}void solve(){    sort(x, x + N);    int lb = 0, ub = INF;    while ((ub - lb) > 1)    {        int mid = (lb + ub) / 2;        if (C(mid)) lb = mid;        else            ub = mid;    }    cout << lb << endl;}int main(){    while (cin >> N >> M)    {        for (int i = 0; i < N; i++)        {            cin >> x[i];        }        solve();    }    return 0;}