牛客多校(2020第八场) K - Kabaleo Lite

参考：https://www.cnblogs.com/lilibuxiangtle/p/13427959.html

题目描述

Tired of boring WFH (work from home), Apollo decided to open a fast food restaurant, called Kabaleo Lite\textbf{Kabaleo Lite}Kabaleo Lite.
The restaurant serves n kinds of food, numbered from 1 to n. The profit for the i-th kind of food is aia_iai​. Profit may be negative because it uses expensive ingredients. On the first day, Apollo prepared bib_ibi​ dishes of the i-th kind of food.
The peculiarity of Apollo's restaurant is the procedure of ordering food. For each visitor Apollo himself chooses a set of dishes that this visitor will receive. When doing so, Apollo is guided by the following rules:

• every visitor should receive at least one dish.
• each visitor should receive continuous kinds of food started from the first food. And the visitor will receive exactly 1 dish for each kind of food. For example, a visitor may receive 1 dish of the 1st kind of food, 1 dish of the 2nd kind of food, 1 dish of the 3rd kind of food.

What's the maximum number of visitors Apollo can feed? And he wants to know the maximum possible profits he can earn to have the maximum of visitors.

输入描述:

The first line of the input gives the number of test case, T\mathbf{T}T (1≤T≤101 \leq \mathbf{T} \leq 101≤T≤10). T\mathbf{T}T test cases follow.
Each test case begins with a line containing one integers n (1≤n≤1051 \le n \le 10^51≤n≤105), representing the number of different kinds of food.
The second line contains n space-separated numbers aia_iai​ (−109≤ai≤109-10^9 \le a_i \le 10^9−109≤ai​≤109), where aia_iai​ denotes the profit of one dish of the i-th kind.
The third line contains n space-separated numbers bib_ibi​ (1≤bi≤1051 \le b_i \le 10^51≤bi​≤105), where bib_ibi​ denotes the number of the i-th kind of dishes.

输出描述:

For each test case, output one line containing ‘‘Case #x: y z′′``Case\ \#x:\ y\ z''‘‘Case #x: y z′′, where x is the test case number (starting from 1), y is the maximum number of visitors, and z is the maximum possible profits.

示例1

输入

2
3
2 -1 3
3 2 1
4
3 -2 3 -1
4 2 1 2

输出

Case #1: 3 8
Case #2: 4 13

说明

For test case 1, the maximum number of visitors is 3, one of a possible solution is:
The first visitor received food 1, the profit is 2.
The second visitor received food 1, the profit is 2.
The third visitor received food 1 + food 2 + food 3, the profit is 2 + (-1) + 3.

题解：

题目大意:

• 有n种菜，每种菜的数量为bi，每个菜的盈利为ai
• 每个顾客必须是从第1种菜开始吃，连续地吃，每种吃一个
• 保证顾客最多的情况下，盈利最大

题解：

• 最大顾客数量就是b1,然后求盈利的前缀和，从大到小取就好(可以使用优先队列protiry_queue)
• 结果会超出long long，可以使用__int128,或者把大数拆开存储(拆成俩个long long)

  1 #include
  2 using namespace std;
  3 
  4 typedef __int128 ll;
  5 const ll N = 1e5 + 5;
  6 ll a[N],b[N];
  7 ll dp[N];//dp存储价钱的前缀和
  8 ll n,t;
  9 
 10 #define P pair> //第一个存dp[i], 第二个存数量, 第三个存菜的下标(第几道菜)
 11 priority_queue 
 que;
 12 
 13 inline int read(){
 14     int x=0,f=1;
 15     char ch=getchar();
 16     while(ch<'0'||ch>'9'){
 17         if(ch=='-')
 18             f=-1;
 19         ch=getchar();
 20     }
 21     while(ch>='0'&&ch<='9'){
 22         x=x*10+ch-'0';
 23         ch=getchar();
 24     }
 25     return x*f;
 26 }
 27 
 28 inline void print(__int128 x){
 29     if(x<0){
 30         putchar('-');
 31         x=-x;
 32     }
 33     if(x>9)
 34         print(x/10);
 35     putchar(x%10+'0');
 36 }
 37 
 38 ll solve() {
 39     ll res = 0;
 40     ll pos_now = n, now_food_num = 0;
 41 
 42     while (!que.empty()) {
 43         ll max_price = que.top().first; //最大利润
 44         ll food_num = que.top().second.first; //菜品数量
 45         ll pos = que.top().second.second; //菜品名
 46         que.pop();
 47         if (pos_now <= pos || food_num <= now_food_num) { //当该菜品位置位于pos_now之后则不可上菜
 48             continue;
 49         }
 50 
 51         res += (food_num - now_food_num) * max_price; //(需要减去上次用的菜品，因为其数量的价值上一次已经算过了)
 52         now_food_num = food_num; //更新上一次用去的菜品数量
 53         pos_now = pos; //更新菜品位置
 54     }
 55     return res;
 56 }
 57 int main() {
 58     /* ios::sync_with_stdio(false);
 59     cin.tie(0); cin.tie(0); */
 60     t = read();
 61     for(ll i = 1; i <= t; i++) {
 62          n = read();
 63         for(ll j = 0; j < n; j++) {
 64             a[j] = read();
 65             if (!j) dp[j] = a[j]; //dp存储价钱的前缀和
 66             else dp[j] = dp[j-1] + a[j];
 67         }
 68 
 69         for(ll j = 0; j < n; j++) {
 70             b[j] = read();
 71         }
 72 
 73         ll bb = b[0];
 74         for (ll j = 0; j < n; j++) {
 75             bb = min(bb, b[j]); //因为当前面的菜没了的时候，后面的菜也相当于没有，所以一直更新前方菜品的最小值
 76             que.push({dp[j], {bb, j}});
 77         }
 78 
 79         ll ans = solve();
 80         cout << "Case #";
 81         print(i);
 82         cout << ": ";
 83         print(b[0]);
 84         cout << " ";
 85         print(ans);
 86         puts("");
 87     }
 88     /* clock_t end = clock();
 89     cout << end - start << "\n"; */
 90     return 0;
 91 }
 92 /* 
 93 2
 94 6
 95 1 1 1 1 1 1
 96 8 3 2 1 9 3
 97 6
 98 2 -1 -3 4 5 -5
 99 10 9 8 7 6 5
100  */