P1824 进击的奶牛（二分）

Posted 2021-02-14 alingmaomao

思路：把检验的函数说一下，就是检测的距离x时，是否存在c个隔断相离大于等于x，如果是则返回1，不是则返回0

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

const int maxn = 1e5 + 10;
int a[maxn], n, c, minn=0x3f3f3f3f, ans, mid;

bool check(int x){
    int sum = 0, base = a[1];
    for (int i = 2; i <= n; ++i){
        if (a[i] - base >= x){ sum++, base = a[i]; }
        if (sum >= c)return 1;
    }
    if (sum + 1 < c)return 0;
    return 1;
}


void half(){
    int l = minn, r = a[n] - a[1];
    while (l <= r){
        mid = (l + r) >> 1;
        if (check(mid)){ l = mid + 1; }
        else r = mid - 1;
    }
    ans = r;
}

int main(){
    scanf("%d%d", &n, &c);
    for (int i = 1; i <= n; ++i)    scanf("%d", &a[i]);
    sort(a + 1, a + n + 1);
    for (int i = 2; i <= n; ++i) minn = min(minn, a[i] - a[i - 1]);
    half();        //二分
    cout << ans << endl;

}