POJ 2456 Aggressive cows 简单二分

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15803   Accepted: 7573 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

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Link：http://poj.org/problem?id=2456

题意：攻击性的牛，为了让牛尽可能不互相攻击，寻找一个牛栏间隔最大距离，能放进所有牛，N为牛栏数，C为牛个数，N行为牛栏所在坐标点

思路：二分法查找牛栏距离，关键在于判断函数

AC code：

#include <cstdio>
#include <algorithm>
using namespace std;


long long a[100050];
long long  mid,cnt;
long long  n,c;
long long  l,r;


int judge(int mid){
    long long  temp=a[0];
    cnt=0;
    for(int i=0;i<n;i++){
        if(a[i]-temp>=mid){
            temp=a[i];
            cnt++;
            //及时返回 否则容易TLE
            if(cnt>=c) return 1;
        }
    }
    return 0;
}


int main()
{
    while(scanf("%lld%lld",&n,&c)!=EOF){
        //个数减一 因为cnt按合法间距累加
        c=c-1;
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
        }
        sort(a,a+n);
        l=a[0];
        r=a[n-1];
        cnt=0;
        //二分框架
        while(l<r){
            //满足即跳出
            if(l+1>=r&&cnt==c) break;
            mid=(l+r)/2;
            if(judge(mid)) l=mid;
            else r=mid;
        }
        printf("%lld\n",mid);
    }
    return 0;
}