poj2456

二分法

有些能用动态规划求解但是超时的题目，可以考虑二分法。

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

#define maxn 100005
#define inf 0x3f3f3f3f

int n, cow;
int pos[maxn];

void input()
{
    scanf("%d%d", &n, &cow);
    for (int i = 0; i < n; i++)
        scanf("%d", &pos[i]);
}

bool ok(int len)
{
    int l = 0, r = 0;
    for (int i = 1; i < cow; i++)
    {
        while (r < n && pos[r] - pos[l] < len)
            r++;
        if (r >= n)
            return false;
        l = r;
    }
    return true;
}

int binary_search()
{
    int l = inf;
    for (int i = 1; i < n; i++)
        l = min(l, pos[i] - pos[i - 1]);
    int r = (pos[n - 1] - pos[0]) / (cow - 1);
    while (l < r)
    {
        int mid = (l + r) / 2 + ((l + r) & 1);
        if (!ok(mid))
            r = mid - 1;
        else
            l = mid;
    }
    return l;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    input();
    sort(pos, pos + n);
    printf("%d\n", binary_search());
    return 0;
}