poj 1952 BUY LOW, BUY LOWER【解法一】

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.

解法一见【这里】。

除了可以用set维护，还可以离散化以后直接用数组存储。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[5010],b[5010],now[5010],dp[5010],f[5010];
bool ok[5010];
int main()
{
int i,j,k,m,n,p,q,x,y,z,max;
scanf("%d",&n);
for (i=1;i<=n;i++)
scanf("%d",&a[i]),b[i]=a[i];
sort(b+1,b+n+1);
m=unique(b+1,b+n+1)-b-1;
for (i=1;i<=n;i++)
now[i]=lower_bound(b+1,b+m+1,a[i])-b;
for (i=1;i<=n+1;i++)
{
memset(ok,1,sizeof(ok));
max=0;
f[i]=1;
for (j=i-1;j;j--)
if (now[j]>now[i])
{
if (dp[j]>max)
{
max=dp[j];
memset(ok,1,sizeof(ok));
ok[now[j]]=0;
f[i]=f[j];
}
if (dp[j]==max&&ok[now[j]])
{
f[i]+=f[j];
ok[now[j]]=0;
}
}
dp[i]=max+1;
}
printf("%d %d\n",dp[n+1]-1,f[n+1]);
}