POJ 1952 Buy Low Buy Lower 最长下降子序列+去重

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title

POJ 1952
CH POJ1952
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

analysis

这题是求最长下降子序列+统计个数，并且计算序列个数时不能重复。设 $f[i]$为以 $a[i]$结尾序列的LIS长度，关键是怎么保证统计个数时不会重复，可增设一个数组 $cnt[i]$，来存储以 $i$为下标的 $f[i]$序列长度的个数。

code

#include<bits/stdc++.h>
using namespace std;
const int maxn=5e4+10;
template<typename T>inline void read(T &x)
{
	x=0;
	T f=1, ch=getchar();
	while (!isdigit(ch) && ch^'-') ch=getchar();
	if (ch=='-') f=-1, ch=getchar();
	while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
	x*=f;
}
int a[maxn],f[maxn],cnt[maxn];
int main()
{
	int n;read(n);
	for (int i=1; i<=n; ++i)
		read(a[i]);
	for (int i=1; i<=n; ++i)
		cnt[i]=f[i]=1;
	for (int i=1; i<=n; ++i)
		for (int j=i-1; j>=0; --j)
			if (a[i]<a[j])
			{
				if (f[i]<f[j]+1)
				{
					f[i]=f[j]+1;
					cnt[i]=cnt[j];
				}
				else if (f[i]==f[j]+1)
					cnt[i]+=cnt[j];
			}
			else if (a[i]==a[j])//去重
			{
				if (f[i]==1)
					cnt[i]=0;
				break;
			}
	int ans=0;
	int num=0;
	for (int i=1; i<=n; ++i)
		ans=max(ans,f[i]);
	for (int i=1; i<=n; ++i)
		if (f[i]==ans)
			num+=cnt[i];
	printf("%d %d\n",ans,num);
	return 0;
}