【贪心算法】POJ-2393 简单贪心水题

一、题目

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.

二、思路&心得

• 这题确实有点水，数据也很弱，比较的应该不只是相邻两周的数据，之后的应该也要比较，但是两种代码都能AC，这就有趣了。。。
• 我的代码比较麻烦点，还有更简单复杂度更小为O(N)的解法，扫描时维护一个min值就行了，每次进入循环时自加S，然后和当前周的Y进行比较取其中最小的即可。

三、代码

解法一：

#include<cstdio>
#include<algorithm>
#define MAX_N 10005
using namespace std;

struct Week {
	int C;
	int Y;
	int visit;
} W[MAX_N];

int N, S;

long long cost;

void solve() {
	cost = 0;
	for (int i = 0; i < N; i ++) {
		scanf("%d %d", &W[i].C, &W[i].Y);
	}
	for (int i = 0; i < N; i ++) {
		if (!W[i].visit) {
			cost += W[i].C * W[i].Y;
			W[i].visit = 1;
			for (int j = i + 1; j < N; j ++) {
				if ((W[i].C + (j - i) * S) <= W[j].C) {
					cost += (W[i].C + (j - i) * S) * W[j].Y;
					W[j].visit = 1;
				} else {
					break;
				}
			}
		}
	}
	printf("%lld
", cost);
}

int main() {
	scanf("%d %d", &N, &S);
	solve();
	return 0;
}

解法二：

#include <cstdio>
typedef long long ll;
struct AC{
	ll c,y;
}r[11000];
ll n,s;
int main(){
	scanf("%d%d",&n,&s);
	for (int i=1;i<=n;i++)
		scanf("%d%d",&r[i].c,&r[i].y);
	ll ans=0,min=1<<30;
	for (int i=1;i<=n;i++){
		min+=s;
		if (min>r[i].c)
			min=r[i].c;
		ans+=min*r[i].y;
	}
	printf("%lld",ans);
	return 0;
}