NC24724 [USACO 2010 Feb S]Chocolate Eating

题目

题目描述

Bessie has received \(N (1 <= N <= 50,000)\) chocolates from the bulls, but doesn't want to eat them too quickly, so she wants to plan out her chocolate eating schedule for the next \(D (1 <= D <= 50,000)\) days in order to maximize her minimum happiness level over the set of those days.
Bessie's happiness level is an integer that starts at 0 and halves (rounding down if necessary) over night as she sleeps. However, when she eats chocolate i, her happiness level increases by integer \(H_i\) (1 <= \(H_i\)​ <= 1,000,000). If she eats chocolates on a day, her happiness for that day is considered the happiness level after she eats the chocolates. Bessie insists that she eat the chocolates in the order that she received them.
If more than one optimal solution exists, print any one of them.
Consider a sequence of 5 chocolates to be eaten over a period of 5 days; they respectively bring happiness (10, 40, 13, 22, 7).

If Bessie eats the first chocolate (10 happiness) on the first day and then waits to eat the others, her happiness level is 10 after the first day.

Here is the complete schedule which turns out to maximize her minimum happiness:
  Day  Wakeup happiness   Happiness from eating   Bedtime happiness
   1            0                10+40                  50
   2           25                 ---                   25
   3           12                  13                   25
   4           12                  22                   34 
   5           17                   7                   24
The minimum bedtime happiness is 24, which turns out to be the best Bessie can do.

输入描述

• Line 1: Two space separated integers: N and D
• Lines 2..N+1: Line i+1 contains a single integer: \(H_i\)

输出描述

• Line 1: A single integer, the highest Bessie's minimum happiness can be over the next D days
• Lines 2..N+1: Line i+1 contains an integer that is the day on which Bessie eats chocolate i

示例1

输入

5 5 
10 
40 
13 
22 
7 

输出

24
1
1
3
4
5

题解

知识点：二分。

二分睡前快乐，起床后快乐不达标就吃巧克力，达标就不管，中间记得记录吃到第几个巧克力。

坑点：最终答案是要把巧克力在最后一刻全吃完，所以如果达标但是巧克力没吃完，记得都输出在最后一天。

时间复杂度 \(O(D)\)

空间复杂度 \(O(D)\)

代码

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int N, D;
int H[50007];
bool flag;
vector<int> ans(50007);

bool check(ll mid) {
    ll h = 0;
    int cnt = 0;
    for (int i = 0;i < D;i++) {
        h >>= 1;
        while (h < mid && cnt < N) {
            h += H[cnt++];
            if (flag) ans[cnt - 1] = i + 1;
        }
        if (h < mid) return false;
    }
    return true;
}

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    cin >> N >> D;
    for (int i = 0;i < N;i++) cin >> H[i];
    ll l = 0, r = 1e12;
    while (l <= r) {
        ll mid = l + r >> 1;
        if (check(mid)) l = mid + 1;
        else r = mid - 1;
    }
    flag = true;
    check(r);
    cout << r << '\n';
    for (int i = 0;i < N;i++) cout << (ans[i] ? ans[i] : D) << '\n';
    return 0;
}