POJ 2456 Aggressive cows （二分查找）

Description

给n个牛栏坐标和c头牛，把c个牛放到n个牛栏上，求满足任意两头牛位置最远的最小值

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.

Soulution

二分法最小化最大值，实现一遍check函数，例如对于数列1 2 4 8 9
，求距离是0～10时，分别能放几头牛，距离是1时能放5头牛，如果大于所给的牛，则距离1太小，需要增加距离即取1到10的mid。

#include <iostream>#include <cmath>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int INF = 1000000000;const int maxn = 110000;int N, C;int x[maxn];bool Check(int mid){    int num = 1, cur = 0, last = 0, i = 0;    while (i < N)    {        if (x[cur] - x[last] >= mid)        {            num++;            last = cur;        }        i++;        cur = i;    }    if (num >= C)    {        return false;    }    else    {        return true;    }}int main(){    while (~scanf("%d%d", &N, &C))    {        memset(x, 0, sizeof(x));        for (int i = 0; i < N; i++)        {            scanf("%d", &x[i]);        }        sort(x, x + N);        int l = 0, r = INF;        for (int t = 0; t < 100; t++)        {            int mid = l + (r - l) / 2;            if (Check(mid))            {                r = mid;            }            else            {                l = mid;            }        }        cout << l << "\n";    }    return 0;}