POJ 1952 BUY LOW, BUY LOWER

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

【题目分析】

动态规划+set

【代码】

#include <set>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
set <int> s;
int maxx,n,a[5010],f[5010],cnt[5010];
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;++i) scanf("%d",&a[i]);
for (int i=1;i<=n+1;++i)
{
s.clear();
maxx=0;
cnt[i]=1;
for (int j=i-1;j;--j)
if (a[i]<a[j])
{
if (f[j]>maxx)
{
maxx=f[j];
s.clear();
s.insert(a[j]);
cnt[i]=cnt[j];
}
else if (f[j]==maxx&&!s.count(a[j]))
{
s.insert(a[j]);
cnt[i]+=cnt[j];
}
}
f[i]=maxx+1;
}
printf("%d %d\n",f[n+1]-1,cnt[n+1]);
}