POJ - 2456 Aggressive cows（二分+贪心）

题意：把c个牛分进n个摊位，摊位位置已知，所有摊位分布在0 <= xi <= 1,000,000,000，问两头牛间最小距离的最大值。

分析：找所有最小距离取个最大的。所以二分找这个最小的距离，这个最大的最小距离是二分的分界线。贪心来判断当前的最小距离是否能安排下所有的牛。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {, , -, , -, -, , };
const int dc[] = {-, , , , -, , -, };
const int MOD = 1e9 + ;
const double pi = acos(-1.0);
const double eps = 1e-;
const int MAXN =  + ;
const int MAXT =  + ;
using namespace std;
int a[MAXN];
int n, c;
bool judge(int x){
    int cnt = ;
    int st = a[];
    for(int i = ; i < n; ++i){
        if(a[i] - st >= x){
            ++cnt;
            st = a[i];
        }
        if(cnt == c) return true;
    }
    return false;
}
int solve(){
    int l = , r = 1e9;
    while(l < r){
        int mid = l + (r - l + ) / ;
        if(judge(mid)) l = mid;
        else r = mid - ;
    }
    return l;
}
int main(){
    while(scanf("%d%d", &n, &c) == ){
        memset(a, , sizeof a);
        for(int i = ; i < n; ++i){
            scanf("%d", &a[i]);
        }
        sort(a, a + n);
        int ans = solve();
        printf("%d\n", ans);
    }
    return ;
}