POJ2393 Yogurt factory 【贪心】

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6821		Accepted: 3488

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. 

Source

USACO 2005 March Gold

每次更新相邻的下一周就可以。由于若下一周被更新，那么下一周能够用来更新剩下的周，所以当前周仅仅须要负责下一周。

#include <stdio.h>
#include <string.h>

#define maxn 10002

int min(int a, int b) {
    return a < b ? a : b;
}

int X[maxn], Y[maxn];

int main() {
    int N, S, i, j;
    __int64 sum;
    while(scanf("%d%d", &N, &S) == 2) {
        for(i = 0; i < N; ++i)
            scanf("%d%d", &X[i], &Y[i]);
        for(i = sum = 0; i < N; ++i) {
            sum += X[i] * Y[i];
            if(i != N - 1)
                X[i+1] = min(X[i+1], X[i] + S);
        }
        printf("%I64d\n", sum);
    }
    return 0;
}

2014-12-1更新

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;

LL ans;

int main() {
    int N, S, i, c, m, pre;
    scanf("%d%d", &N, &S);
    scanf("%d%d", &c, &m);
    pre = c + S; // 仓库价格
    ans += c * m;
    while(--N) {
        scanf("%d%d", &c, &m);
        if(pre < c) c = pre;
        else pre = c;
        ans += c * m;
        pre += S;
    }
    printf("%lld\n", ans);
    return 0;
}