poj 2456 Aggressive cows，二分，最大化最小值

3

排序后，二分最小距离

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

const int maxn = 100000 + 10;

int x[maxn];
int n, c;

int ok(int k) {
    int last = 0;
    int cur;
    for(int i=1; i<c; ++i) {
        cur = last + 1;
        while(cur<n && x[cur] - x[last] < k)
            cur++;
        if(cur==n) return false;
        last = cur;
    }
    return true;
}
int main() {
    scanf("%d%d",&n, &c);
    for(int i=0; i<n; ++i) scanf("%d", &x[i]);
    sort(x, x+n);

    int l = 0, r = (x[n-1] - x[0])/(c-1);
    for(int i=0; i<100; ++i) {
        int mid = (l+r)/ 2;
        if( !ok(mid) )  r = mid;
        else l = mid;
    }
    printf("%d\n", l);
    return 0;
}

poj 2456 Aggressive cows，二分，最大化最小值,布布扣,bubuko.com

poj 2456 Aggressive cows，二分，最大化最小值

原文：http://blog.csdn.net/yew1eb/article/details/38725479