2013年长沙区域赛-HDU 4791-Alice's Print Service

最新推荐文章于 2018-10-15 00:11:16 发布

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#思维 #STL

本文介绍了一种基于分段定价的打印服务策略，通过算法找出在不同页数区间内的最优打印成本，实现用户节省费用的目的。

Alice's Print Service

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2230 Accepted Submission(s): 585

Problem Description

Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.
For example, the price when printing less than 100 pages is 20 cents per page, but when printing not less than 100 pages, you just need to pay only 10 cents per page. It's easy to figure out that if you want to print 99 pages, the best choice is to print an extra blank page so that the money you need to pay is 100 × 10 cents instead of 99 × 20 cents.
Now given the description of pricing strategy and some queries, your task is to figure out the best ways to complete those queries in order to save money.

Input

The first line contains an integer T (≈ 10) which is the number of test cases. Then T cases follow.
Each case contains 3 lines. The first line contains two integers n, m (0 < n, m ≤ 10⁵ ). The second line contains 2n integers s₁, p₁ , s₂, p₂ , ..., s_n, p_n (0=s₁ < s₂ < ... < s_n ≤ 10⁹ , 10⁹ ≥ p₁ ≥ p₂ ≥ ... ≥ p_n ≥ 0).. The price when printing no less than s_i but less than s_i+1 pages is p_i cents per page (for i=1..n-1). The price when printing no less than s_n pages is p_n cents per page. The third line containing m integers q₁ .. q_m (0 ≤ q_i ≤ 10⁹ ) are the queries.

Output

For each query q_i, you should output the minimum amount of money (in cents) to pay if you want to print q_i pages, one output in one line.

Sample Input

  
   1
2 3
0 20 100 10
0 99 100

Sample Output

Source

2013 Asia Changsha Regional Contest

题意：按纸张数量分段，价格不同，给出若干次询问，求印费最少为多少？

解题思路：计算出当前价格和印更多张价格的最小值，两者相互比较，取较小者。

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;

LL s[100005],p[100005];
LL sp[100005];
LL dp[100005];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    int q;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            cin>>s[i]>>p[i];
            sp[i]=s[i]*p[i];
        }
        dp[n-1]=sp[n];
        for(int i=n-2;i>=1;i--)
        {
            if(sp[i+1]<dp[i+1]) dp[i]=sp[i+1];
            else dp[i]=dp[i+1];
        }
        for(int i=1;i<=m;i++)
        {
            cin>>q;
            LL ts1=lower_bound(s+1,s+n+1,q)-(s+1);
            LL ts2=ts1;
            if(q==s[ts1+1]) ts1++;
            LL nowprice=q*p[ts1];
            if(ts1==n)
                cout<<nowprice<<endl;
            else
            {
                if(nowprice>dp[ts2]) cout<<dp[ts2]<<endl;
                else cout<<nowprice<<endl;
            }
        }
    }
    return 0;
}