2590: [Usaco2012 Feb]Cow Coupons

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 306 Solved: 154
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Description

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? PROBLEM NAME: coupons

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？

Input

* 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.

Output

* Line 1: A single integer, the maximum number of cows FJ can afford.

Sample Input

4 1 7
3 2
2 2
8 1
4 3

Sample Output

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

跟着大爷跑来写了波题找个时间也去刷刷USACO（觉得要跪）

这道题据说网上大部分题解都是错的？？？手动黑人问号脸可能数据比较水吧

据tjm大爷讲：

首先肯定选k个优惠券最小的，如果不够选直接输出。如果还有多余的钱的话，比较一下（已买的里面原价与优惠价的最小差价+没买的里面的最小优惠价）和（没买的里面的最小原价），选一个小的。第一个实际上相当于把优惠卷花钱赎回来。。。神犇传送门（tjm大爷的博客）

我的思路自然是照着大爷写的一波写法可能有点小不同但思路一样就好啦 2333

当然还得不要脸地贴一波自己的博客啦

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue> 
#define LL long long
using namespace std;
const int M=55007;
const LL inf=1e15;
LL read(){
    LL ans=0,f=1,c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
    return ans*f;
}
LL m,n,k,v,cost;
LL w[M],p[M],ans,sum;
bool f[M];
struct node{
    LL w,pos;
    bool operator < (const node& x)const {return x.w<w;}
};
priority_queue<node>q1,q2,q3;
int main()
{
    n=read(); k=read(); m=read();
    k=min(k,n);
    for(int i=1;i<=n;i++){
        w[i]=read(); q1.push((node){w[i],i});
        p[i]=read(); q2.push((node){p[i],i});
    }
    for(ans=1;ans<=k;ans++){
        node x=q2.top();
        if(cost+x.w>m){printf("%lld
",ans-1); return 0;}
        q2.pop(); f[x.pos]=1;
        cost+=x.w;
        q3.push((node){w[x.pos]-p[x.pos],x.pos});
    }ans--;
    if(ans==n){printf("%lld
",ans); return 0;}
    while(cost<=m&&ans<n){
        node x=q3.top(),y=q2.top(),z=q1.top();
        while(f[y.pos]&&!q2.empty()) q2.pop(),y=q2.top();
        if(q2.empty()) y.w=inf;
        while(f[z.pos]&&!q1.empty()) q1.pop(),z=q1.top();
        if(q1.empty()) z.w=inf;
        if(x.w+y.w<=z.w) f[y.pos]=1,q3.pop(),q2.pop(),q3.push((node){w[y.pos]-p[y.pos],y.pos}),cost+=x.w+y.w ;
        else q1.pop(),f[z.pos]=1,cost+=z.w;
        if(cost<=m) ans++;
    }
    printf("%lld
",ans);
    return 0;
}

View Code