POJ 1952 BUY LOW, BUY LOWER 动态规划题解

Description

The advice to "buy low" is half the formula to success in the bovine stock market.To be considered a great investor you must also follow this problems' advice: 

                    "Buy low; buy lower"

Each time you buy a stock, you must purchase it at a lower price than the previous time you bought it. The more times you buy at a lower price than before, the better! Your goal is to see how many times you can continue purchasing at ever lower prices. 

You will be given the daily selling prices of a stock (positive 16-bit integers) over a period of time. You can choose to buy stock on any of the days. Each time you choose to buy, the price must be strictly lower than the previous time you bought stock. Write a program which identifies which days you should buy stock in order to maximize the number of times you buy. 

Here is a list of stock prices: 

 Day   1  2  3  4  5  6  7  8  9 10 11 12

Price 68 69 54 64 68 64 70 67 78 62 98 87

The best investor (by this problem, anyway) can buy at most four times if each purchase is lower then the previous purchase. One four day sequence (there might be others) of acceptable buys is: 

Day    2  5  6 10

Price 69 68 64 62

Input

* Line 1: N (1 <= N <= 5000), the number of days for which stock prices are given 

* Lines 2..etc: A series of N space-separated integers, ten per line except the final line which might have fewer integers. 

Output

Two integers on a single line: 
* The length of the longest sequence of decreasing prices 
* The number of sequences that have this length (guaranteed to fit in 31 bits) 

In counting the number of solutions, two potential solutions are considered the same (and would only count as one solution) if they repeat the same string of decreasing prices, that is, if they "look the same" when the successive prices are compared. Thus, two different sequence of "buy" days could produce the same string of decreasing prices and be counted as only a single solution. 

Sample Input

12
68 69 54 64 68 64 70 67 78 62
98 87

Sample Output

4 2

Source

USACO 2002 February

求递减子序列的题目，不过本题多了一个要求，需要统计这样的最长递减子序列的个数，并且需要去掉重复的子序列。

思路就是需要统计当前下标为结束的时候的最长子序列，然后求这个子序列的数，还需要和前面一样的子序列去重。

详细注释的代码：

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <limits.h>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;

const int MAX_N = 5001;
int arr[MAX_N];//数据记录
int C[MAX_N];//C[i]，下标为i的时候最长子序列有多少个
int MaxLen[MAX_N];//MaxLen[i]，下标为i的时候，最长子序列多长

void getLenAndNum(int &len, int &c, int n)
{
	memset(C, 0, sizeof(int) * n);
	C[0] = MaxLen[0] = 1;
	for (int i = 1; i < n; i++)
	{
		MaxLen[i] = 1;//初始值为只有一个
		for (int j = 0; j < i; j++)
		{
			if (arr[j] > arr[i] && MaxLen[i] < MaxLen[j] + 1)
				MaxLen[i] = MaxLen[j] + 1;
		}//求以当前i下标结束的时候，最长子序列长度
		for (int j = 0; j < i; j++)
		{
			if (arr[j] > arr[i] && MaxLen[i] == MaxLen[j] + 1)
				C[i] += C[j];
		}//求有多少最长子序列
		if (!C[i]) C[i] = 1;//注意是递增数列的时候
		for (int j = 0; j < i; j++)
		{
			if (arr[j] == arr[i] && MaxLen[i] == MaxLen[j]) C[i] -= C[j];
		}//去掉重复计算，方便后面的统计
	}
	len = 0;
	for (int i = 0; i < n; i++)
	{
		if (len < MaxLen[i]) len = MaxLen[i];
	}//找出最长子序列
	c = 0;
	for (int i = 0; i < n; i++)
	{
		if (len == MaxLen[i]) c += C[i];
	}//找出最长子序列数
}

int main()
{
	int N;
	while (~scanf("%d", &N))
	{
		for (int i = 0; i < N; i++)
		{
			scanf("%d", arr+i);
		}
		int len, c;
		getLenAndNum(len, c, N);
		printf("%d %d\n", len, c);
	}
	return 0;
}