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]);}