HDU4791——Alice's Print Service(线段树)
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Alice's Print Service
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2240 Accepted Submission(s): 589
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.
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 , ..., 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.
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.
Output
For each query qi, you should 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
Source
2013 Asia Changsha Regional Contest
题意:给你n-1个区间,每个区间有个价格,m个询问,查询打印这些纸张得最小价格。可以打印空白得纸张。
线段树水题,首先查询纸张属于得区间,那么最小值只有可能是,纸张乘以其本身纸张价格,或者之后得每个区间头乘以其区间价格。这样就是一个RMQ问题了。
#include <cmath>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <algorithm>#include <string>#include <map>#include <queue>#include <set>#include <stack>#include <ctime>#include <cstdlib>#include <iostream>#define pi 3.1415926#define mod 1000000007#define MAXN 100010#define INF 0x3f3f3f3f3f3f3fLLusing namespace std;long long s[MAXN],p[MAXN];#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1long long MIN[MAXN<<2];void pushup(int rt){ MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);}void build(int l,int r,int rt){ if(l==r){ MIN[rt]=INF; return; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt);}void update(int p,long long x,int l,int r,int rt){ if(l==r){ MIN[rt]=x; return; } int m=(l+r)>>1; if(p<=m) update(p,x,lson); else update(p,x,rson); pushup(rt);}long long query(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R){ return MIN[rt]; } int m=(l+r)>>1; long long res=INF; if(L<=m){ res=min(res,query(L,R,lson)); } if(R>m){ res=min(res,query(L,R,rson)); } return res;}int main(){ int t,n,m; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); build(1,n,1); for(int i=1;i<=n;i++){ scanf("%lld%lld",s+i,p+i); long long temp=s[i]*p[i]; update(i,temp,1,n,1); } while(m--){ int q; scanf("%d",&q); long long ans; if(q>=s[n]) ans=q*p[n]; else if(q==0) ans=0; else{ int l=1,r=n; while(l<r) { int m=(l+r+1)/2; if(s[m]<=q){ l=m; } else r=m-1; // cout<<l<<" "<<r<<endl; } //cout<<l<<" "<<r<<endl; ans=min(q*p[l],query(l+1,n,1,n,1)); } printf("%lld\n",ans); } }}
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