POJ2393————Yogurt factory (贪心)

题目如下：

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9225		Accepted: 4695

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. 

大意：求做酸奶的最小费用。每周都有不同的成本和需求量。这周生产的酸奶可以储存下来给以后的周次，但每保存一周需要S的费用。

思路：简单的贪心算法。如果保存费用加上原来的成本小于这周的成本，原来就多生产一部分保存到这周来。至于为什么是正确的，简单推算一下就知道，如果(c[i]+s*(j-i))<c[j]，那么到第j+1周的时候，(c[i]+s*(j-i)+s)<(c[j]+s)，一个很简单的逻辑。顺着推即可。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>
#include <cctype>
#include <string>
#include <map>
#include <set>
#include <iostream>
using namespace std;

#define INF 1e9+7
#define MAXN 10010

int n,s;
int c[MAXN],y[MAXN];

int main()
{
scanf("%d%d",&n,&s);
for(int i=0;i<n;i++)
scanf("%d%d",c+i,y+i);
long long res=0;
int i=0;
while(i<n)
{
res+=c[i]*y[i];
int j=i+1;
while(j<n)
{
if((s*(j-i)+c[i])<c[j])
{
res+=(c[i]+s*(j-i))*y[j];
j++;
}
else
break;
}
i=j;
}
printf("%lld\n",res);
}