Yogurt factory-POJ2393

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week. 

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt. 

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S. 

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 588 20089 40097 30091 500

Sample Output

126900

Hint

OUTPUT DETAILS: 
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units. 

题解：若考虑到第i个时，就要在前0-i个里面选取一个价格，考虑到保存费，0-i中第k个存价格时就用C[k]-s*k来存，取出0-i里面最小的，此处用堆就好了。

#include"stdio.h"#include"string.h"#include"cstdio"#include"algorithm"#include"queue"using namespace std;const int max_n=1e4+10;typedef long long ll;ll C[max_n],Y[max_n];ll num[max_n];struct Fac{ll cost,sub;friend bool operator<(const Fac &a,const Fac &b){return a.cost>b.cost;}};int main(){int n,s;while(scanf("%d%d",&n,&s)!=EOF){int i;priority_queue<Fac> q;Fac pp;for(i=0;i<n;i++)scanf("%lld%lld",&C[i],&Y[i]);for(i=0;i<n;i++){pp.cost=C[i]-i*s;pp.sub=i;q.push(pp);pp=q.top();num[i]=pp.sub;//printf("%lld\n",pp.sub);}ll ret=0;for(i=0;i<n;i++)ret+=(Y[i]*C[num[i]]+(i-num[i])*s*Y[i]);printf("%lld\n",ret);}}