HDU 4791 Alice's Print Service (DP + offline processing)

Alice's Print Service Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 1163 accepted submission (s): 270


Problem descriptionalice 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.
Inputthe 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 ,..., SN, Pn (0 = S1 <S2 <... <Sn ≤ 109,109 ≥ P1 ≥ P2 ≥... ≥ PN ≥ 0 ).. the price when printing no less than SI but less than SI + 1 pages is pi cents per page (for I = 1 .. n-1 ). the price when printing no less than Sn pages is pn cents per page. the third line containing M integers Q1 .. QM (0 ≤ Qi ≤109) are the queries.
Outputfor each query Qi, you shocould output the minimum amount of money (in cents) to pay if you want to print Qi pages, one output in one line.
Sample Input

12 30 20 100 100 99 100

 
Sample output

010001000

 

Question: For N solutions, when the print volume is greater than or equal to Si, the printing price for each piece of paper is Pi (0 ≤ n ≤ 100000 ).

Ask the minimum cost required for printing a Qi image (0 ≤ m ≤ 100000 ).

Train of Thought: store all the inquiries, sort them in ascending order of requirements, find the answers in sequence, and then output them.

#include <iostream>#include <algorithm>#include <cstdio>#define ll long longusing namespace std;const int maxn=100005;struct node{    ll s,p,cost;}a[maxn];struct code{    int index;    ll ans,num;}b[maxn];int n,m;bool cmp1(code p,code q){    return p.num<q.num;}bool cmp2(code p,code q){    return p.index<q.index;}void input(){    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)  scanf("%I64d%I64d",&a[i].s,&a[i].p);    for(int i=0;i<m;i++)    {         scanf("%I64d",&b[i].num);         b[i].index=i+1;    }    sort(b,b+m,cmp1);}void solve(){    a[n-1].cost=a[n-1].s*a[n-1].p;    for(int i=n-2;i>=0;i--)  a[i].cost=min(a[i].s*a[i].p,a[i+1].cost);    int cnt=0;    for(int i=0;i<m;i++)    {         while(cnt<n && a[cnt].s<=b[i].num)  cnt++;         if(cnt==n)  b[i].ans=b[i].num*a[n-1].p;         else  b[i].ans=min(b[i].num*a[cnt-1].p,a[cnt].cost);    }    sort(b,b+m,cmp2);    for(int i=0;i<m;i++)   printf("%I64d\n",b[i].ans);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        input();        solve();    }    return 0;}

HDU 4791 Alice's Print Service (DP + offline processing)