NYOJ 208 Supermarket （贪心 && 并查集）

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Supermarket

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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)=Σ_x∈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. 

输入

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.

输出

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.

样例输入

4  50 2  10 1   20 2   30 17  20 1   2 1   10 3  100 2   8 2   5 20  50 10

样例输出

80185

这个题归到数据结构应该是使用并查集做，但是可以使用贪心直接做出来。

本来是以为去每个日期的最大值，其实不是，正确的贪心策略是先从最值钱的货物开始售卖，如果当前的货物所在deadline已经

出售了东西，那就在1~deadline之间查找哪天能卖出去。

 #include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 10000 + 10;struct P{    int x, y;     //x是价值，  y是天数}p[maxn];int n;bool vis[maxn];bool cmp(P a, P b){    return a.x > b.x;}int main(){    while (scanf("%d", &n) != EOF){        memset(vis, 0, sizeof(vis));        for (int i = 0; i < n; i++)            scanf("%d%d", &p[i].x, &p[i].y);        sort(p, p + n, cmp);        int sum = 0;        for (int i = 0; i < n; i++){            for (int j = p[i].y; j > 0; j--){                if (!vis[j]){                    sum += p[i].x;                    vis[j] = 1;                    break;                }            }        }        printf("%d\n", sum);    }    return 0;}        

使用并查集

 #include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 10000 + 10;struct P{    int x, y;     //x是价值，  y是天数}p[maxn];int n;int father[maxn];bool cmp(P a, P b){    return a.x > b.x;}int find(int x){    if (x != father[x])        father[x] = find(father[x]);    return father[x];}int main(){    while (scanf("%d", &n) != EOF){        for (int i = 0; i < maxn; i++)            father[i] = i;        for (int i = 0; i < n; i++)            scanf("%d%d", &p[i].x, &p[i].y);        sort(p, p + n, cmp);        int sum = 0;        for (int i = 0; i < n; i++){            int ans = find(father[p[i].y]);            if (ans > 0){                sum += p[i].x;                father[ans] = ans - 1;            }        }        printf("%d\n", sum);    }    return 0;}