poj 2393 Yogurt factory

题目链接
Yogurt factory
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10470 Accepted: 5307
Description

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 5
88 200
89 400
97 300
91 500
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.

#include<iostream>
#include<cstdio>

using namespace std;
typedef long long ll;
struct node{
    ll p,num,id;
}x[10005];
ll slove(ll n,ll s){
    ll ans=x[0].p*x[0].num , kase = 0;
    for(int i = 1 ; i<n ; i++){
        ll tp = x[kase].p+s*(i-x[kase].id),cur=x[i].p;
         if(tp<cur){

            ans+=tp*x[i].num;
        }
        else{

            ans+=cur*x[i].num;
            kase=i;
        }
    }
    return ans;
}
int main(){

    ll n,s;
    while(~scanf("%lld%lld",&n,&s)){
        for(ll i=0;i<n;i++){
            scanf("%lld%lld", &x[i].p,&x[i].num);
            x[i].id=i;
        }
        printf("%lld\n",slove(n,s));
    }
    return 0;
}