Supermarket
A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σx∈Sellpx. An optimal selling schedule is a schedule with a maximum profit. For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80. Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.
Input
A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.
Output
For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.
Sample Input
4 50 2 10 1 20 2 30 1
7 20 1 2 1 10 3 100 2 8 2 5 20 50 10
Sample Output
80
185
Hint
The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.
Source
题目类型：贪心
算法分析：将输入的每对数据按照价值递减排序，然后从前到后依次判断当前商品的最后期限，使用一个bool数组vis标记某一天是否已经被占用过了。如果当前的商品的最后期限没有被占用，则直接在该天售出。否则向前寻找最近没有被标记过的日期。如果有的话，直接在该天售出。贪心策略是：出售价值尽可能大的商品，如果要出售这个商品，要将该物品尽量放在后面出售
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-6; const double PI = 2 * acos (0.0); const int maxn = 1e4 + 66; struct node { int p, d; }; node aa[maxn]; bool vis[maxn]; bool cmp (node a, node b) { return a.p > b.p; } int main() { // CPPFF; int n; while (cin >> n) { for (int i = 1; i <= n; i++) cin >> aa[i].p >> aa[i].d; sort (aa + 1, aa + 1 + n, cmp); LL res = 0; memset (vis, false, sizeof (vis)); for (int i = 1; i <= n; i++) { if (!vis[aa[i].d]) { vis[aa[i].d] = true; res += aa[i].p; } else { for (int j = aa[i].d - 1; j >= 1; j--) { if (!vis[j]) { vis[j] = true; res += aa[i].p; break; } } } } cout << res << endl; } return 0; } |
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