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  • 洛谷P3045 [USACO12FEB]牛券Cow Coupons

    P3045 [USACO12FEB]牛券Cow Coupons

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      • 248提交
    • 题目提供者洛谷OnlineJudge
    • 标签USACO2012云端
    • 难度提高+/省选-
    • 时空限制1s / 128MB

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    • 86分求救

    题目描述

    Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his budget of M units of money (1 <= M <= 10^14). Cow i costs P_i money (1 <= P_i <= 10^9), but FJ has K coupons (1 <= K <= N), and when he uses a coupon on cow i, the cow costs C_i instead (1 <= C_i <= P_i). FJ can only use one coupon per cow, of course.

    What is the maximum number of cows FJ can afford?

    FJ准备买一些新奶牛,市场上有N头奶牛(1<=N<=50000),第i头奶牛价格为Pi(1<=Pi<=10^9)。FJ有K张优惠券,使用优惠券购买第i头奶牛时价格会降为Ci(1<=Ci<=Pi),每头奶牛只能使用一次优惠券。FJ想知道花不超过M(1<=M<=10^14)的钱最多可以买多少奶牛?

    输入输出格式

    输入格式:
    • Line 1: Three space-separated integers: N, K, and M.

    • Lines 2..N+1: Line i+1 contains two integers: P_i and C_i.
    输出格式:
    • Line 1: A single integer, the maximum number of cows FJ can afford.

    输入输出样例

    输入样例#1:
    4 1 7 
    3 2 
    2 2 
    8 1 
    4 3 
    
    输出样例#1:
    3 
    

    说明

    FJ has 4 cows, 1 coupon, and a budget of 7.

    FJ uses the coupon on cow 3 and buys cows 1, 2, and 3, for a total cost of 3 + 2 + 1 = 6.

    分析:其实很容易发现这就是一道背包题,对于每头牛我们都有用与不用优惠券两种选择,然而会发现,这个m不是一般的大,所以不能用dp.dp和贪心是差不多的,考虑到dp不行,试试贪心。因为我们的目标是要使买的牛最多,也就是花的钱最少,于是我当时想了一种贪心:我们可以取前k个用优惠券的价格(从小到大排序),然后和不排序的放在一起排序一下,然后遍历求解.这样的话有一个问题:我们已经假定前k个用优惠券的牛用优惠券,然而有时候不用优惠券比用优惠券要好,那就是用不用价格都相等的情况,所以我们不再取前k个,我们把每头牛拆成2头牛,一头用优惠券,一头不用,然后排序求解即可.

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <functional>
    
    using namespace std;
    
    int n, k,p[50010],c[50010],vis[50010],ans;
    long long m;
    
    struct node
    {
        int id, use, money;
    }e[100010];
    
    bool cmp(node a, node b)
    {
        if (a.money == b.money)
            return a.use < b.use;
        return a.money < b.money;
    }
    
    int main()
    {
        scanf("%d%d%lld", &n, &k, &m);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d", &p[i], &c[i]);
            e[i * 2 - 1].id = i;
            e[i * 2 - 1].use = 1;
            e[i * 2 - 1].money = c[i];
    
            e[i * 2].id = i;
            e[i * 2].use = 0;
            e[i * 2].money = p[i];
        }
        sort(e + 1, e + n * 2 + 1, cmp);
        for (int i = 1; i <= n * 2; i++)
        {
            if (vis[e[i].id])
                continue;
            if (e[i].use && k <= 0)
                continue;
            if (m <= 0)
                break;
            if (m >= e[i].money)
            {
                vis[e[i].id] = 1;
                ans++;
                m -= e[i].money;
                if (e[i].use)
                    k--;
            }
        }
    
        printf("%d", ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zbtrs/p/7071526.html
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