POJ 2393 Yogurt factory（贪心）

总时间限制: 1000ms 内存限制: 65536kB

描述

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.

输入

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.

输出

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.

样例输入

4 5
88 200
89 400
97 300
91 500

样例输出

126900

提示

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.

一开始的代码是这样的，复杂度为O(n^2)

#include <cstring>
#include <iostream>
#include <queue>
#include <math.h>
#include <vector>
#include <algorithm>
#include <map>#define MAX_N 1010
#define MAX_M 10010
using namespace std;int main()
{int n, s;int Y[MAX_M] = {0}, C[MAX_M] = {0};long long total = 0;long long tmpmin = 0x3f3f3f;while (scanf("%d%d", &n, &s) != EOF) {memset(Y, 0, sizeof(Y));memset(C, 0, sizeof(C));total = 0;for (int i = 0; i < n; i++) {scanf("%d%d", &C[i], &Y[i]);}for (int i = n - 1; i >= 0; i--) {tmpmin = 1ll * C[i] * Y[i];for (int j = 0; j < i; j++) {long long curval = 1ll * C[j] * Y[i] + 1ll * (i-j) * s * Y[i];if (curval < tmpmin) {tmpmin = curval;}}total += tmpmin;}printf("%lld\n", total);}return 0;
}

后来看到有其他的做法，复杂度只有O(n)，

#include <cstring>
#include <iostream>
#include <queue>
#include <math.h>
#include <vector>
#include <algorithm>
#include <map>#define MAX_N 1010
#define MAX_M 10010
using namespace std;int main()
{int n, s, y, c, m;long long total = 0;while (scanf("%d%d", &n, &s) != EOF) {total = 0;m = 0x3f3f3f;while (n--) {scanf("%d%d", &c, &y);m = min(m+s, c);total += 1ll * m * y;}printf("%lld\n", total);}return 0;
}

区别就在于如何遍历所有的日期，我的做法是从后面开始遍历，对于每一天，再从第一天一直找到这一天期间所有的价格，找到一个最小的加上，这样复杂度很高。第二种方法使用m=min(m+s,c)通过将单价c与前一日单价与保存单价之间的比较来得到最后的结果，效率更高。

不过一定要注意的是，对于中间计算结果一定要乘上1ll，就是1后面加一个longlong转换类型，不然会疯狂WA。