贪心--poj2393

Language: Default
Yogurt factory
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5634   Accepted: 2898

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

很简单的贪心,有让自己整复杂了。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int N,S;
int c,y;
int min1=1000000000;
int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>N>>S)
    {
        long long sum=0;
        for(int i=0;i>c>>y;
            if(c>min1+S) c=min1+S;
            min1=c;
            sum+=c*y;
        }
        cout<


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