POJ 2456 Aggressive cows（二分）

Aggressive cows

题目链接：

http://poj.org/problem?id=2456

解题思路：

1. 对牛舍的位置x进行排序
2. 把第一头牛放入x0的牛舍
3. 如果把i头牛放入了xj的话，第i+1头牛就要放入满足xj+d<=xk的最小的xk中

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#define INF 0xfffffffusing namespace std;int n,m;int a[100005];bool check(int d){    int last = 0;    for(int i = 1; i < m; i++){        int cur = last + 1;        while(cur < n && a[cur] - a[last] < d)            cur++;        if(cur == n)            return false;        last = cur;    }    return true;}int main(){    while(~scanf("%d%d",&n,&m)){        for(int i = 0; i < n; i++)            scanf("%d",&a[i]);        sort(a,a+n);        int l = 0, r = INF;        while((r-l) > 1){            int mid = (l+r)>>1;            if(check(mid))                l = mid;            else                r = mid;        }        printf("%d\n",l);    }    return 0;}