牛券Cow Coupons

USACO12FEB

久违的奶牛题。

题意:

FJ准备买一些新奶牛,市场上有 $ N $ 头奶牛 $ (1 \leq N \leq 50000) $ ,第i头奶牛价格为 $ P_i (1 \leq P_i \leq 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 $ C_i(1\leq C_i \leq P_i) $ ,每头奶牛只能使用一次优惠券。FJ想知道花不超过 $ M(1 \leq M \leq 10^{14}) $ 的钱最多可以买多少奶牛?

解法:

在ZR时摸鱼王讲的一道贪心题。
但这道题并不是一道裸贪心,直接对 $ C $ 排序,取前 $ k $ 个数并不完全对,具体为什么自己想想。
正确的做法依旧是贪心,不过是可以反悔的贪心。
我们优先处理使用优惠券之后最便宜的几头牛,然后选择剩下的牛中不用券最便宜的,之后判断要不要将用过的一张券转用给一头新的牛。
具体做法就是开一个大根堆,每次维护 $ price_i - cost_i $ 就可以了。

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;

#define LL long long
#define N 50010

LL n,k,m;

struct cow {
    LL price,cost;
} node[N];
inline bool cmp1(cow a,cow b) {
    return a.cost < b.cost;
}
inline bool cmp2(cow a,cow b) {
    return a.price < b.price;
}
priority_queue<LL,vector<LL>,greater<LL> > q;

int main() {
    scanf("%lld%lld%lld",&n,&k,&m);
    for(int i = 1 ; i <= n ; i++)
        scanf("%lld%lld",&node[i].price,&node[i].cost);
    sort(node + 1,node + n + 1,cmp1);
    LL sum = 0;
    for(int i = 1 ; i <= k ; i++) {
        sum += node[i].cost;
        if(sum > m) {
            printf("%d \n",i - 1);
            //system("pause");
            return 0;
        }
        q.push(node[i].price - node[i].cost);
    }
    sort(node + k + 1,node + n + 1,cmp2);
    for(int i = k + 1 ; i <= n ; i++) {
        int u = node[i].price - node[i].cost;
        if(u > q.top()) {
            sum += q.top();
            q.pop();
            q.push(u);
            sum += node[i].cost;
        }
        else sum += node[i].price;
        if(sum > m) {
            printf("%d \n",i - 1);
            //system("pause");
            return 0;
        }
    }
    printf("%lld \n",n);
    //system("pause");
    return 0;
}

原文地址:https://www.cnblogs.com/Repulser/p/11402679.html

时间: 2024-11-02 16:54:44

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