[USACO12FEB]牛券Cow Coupons

最新推荐文章于 2021-08-12 13:29:34 发布

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最新推荐文章于 2021-08-12 13:29:34 发布 · 569 阅读

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本文讨论了USACO竞赛中的一道问题，FJ需要在预算M内购买N头奶牛，每头奶牛原价P_i，拥有K张优惠券可将价格降至C_i。目标是找出FJ最多能买多少头奶牛。输入包括奶牛数量、优惠券数量和预算，输出最大购买数。示例展示了当FJ有4头奶牛、1张优惠券和7单位预算时，通过优惠券购买策略，他可以买下3头奶牛。

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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：复制

4 1 7
3 2
2 2
8 1
4 3

输出样例#1：复制

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.

解法一：60分
用一个堆排序，并对用过的点打标记
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
using namespace std;

int n,k,p,c,ans;
long long m;
bool bo[50005];
struct cyq{
   
   
    int mo,no;
    int pa;
}d;
priority_queue <cyq,vector<cyq>,greater<cyq> > q;
bool operator >(const cyq &aa,const cyq &bb){
   
   
    return aa.mo>bb.mo;
}

int main()
{
   
   
    scanf("%d%d%lld",&n,&k,&m);
    for(int i=1;i<=n;i++){
   
   scanf("%d%d",&p,&c);q.push((cyq){
   
   p,i,0