POJ 1952 BUY LOW, BUY LOWER

BUY LOW, BUY LOWER

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8311		Accepted: 2883

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

哎，做这道题目的时候天知道我都经历了一些什么，最终过了，哭的心都有。

题意：求最长的下降序列的个数

想法：对于最长的下降序列能够用nlog(n)的算法。然后就是构图。x->y，找出哪些点x能到y,满足假设rank(y) = a,

则rank(x) = a-1; val(x)<val(y);id(x)<id(y),假设x点不同，那直接加边就好了，假设x点有同样的。取近期的那个。这样就不会有重算的，由于近期的那个肯定包括了最远的在里面了

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <map>
#define N 5100
#define M 800010
#define INF 0x7ffffff
using namespace std;
struct num
{
    int id,next;
}a[M],b[M];
int c[N],d[N],sum[N],tag[N];
int e[N],f[N],Top1,Top2;
bool ch[N],ch2[N];
map<int,int>ma;
int main()
{
    //freopen("data.txt","r",stdin);
    void addeage1(int x,int y);
    void addeage2(int x,int y);
    int dfs(int x,int pt);
    int binary_search(int l,int r,int val);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        ma.clear();
        for(int i=n;i>=1;i--)
        {
            scanf("%d",&e[i]);
            ma[e[i]] = i;
        }
        e[n+1] = INF;
        ma[INF] = n+1;
        memset(c,-1,sizeof(c));
        Top1 = 0;
        memset(d,-1,sizeof(d));
        Top2 = 0;
        f[0] = -1;
        int Top = 0;
        addeage1(n+1,n+1);
        for(int i=1;i<=n;i++)
        {
            if(f[Top]<e[i])
            {
                f[++Top] = e[i];
                addeage1(Top,i);
            }else
            {
                int pos = binary_search(1,Top,e[i]); //第一个大于||等于
                f[pos] = e[i];
                addeage1(pos,i);
            }
        }
        int biao = 0;
        sum[biao++] = n+1;
        for(int i=Top;i>=1;i--)
        {
            int ttag = 0;
            memset(ch2,false,sizeof(ch2));
            for(int j=0;j<=biao-1;j++)
            {
                memset(ch,false,sizeof(ch));
                for(int u = c[i];u!=-1;u=a[u].next)
                {
                    int id = a[u].id;
                    if(id<sum[j]&&e[id]<e[sum[j]]&&!ch[ma[e[id]]])
                    {
                        ch[ma[e[id]]] = true;
                        addeage2(sum[j],id);
                        if(!ch2[id])
                        {
                            tag[ttag++] = id;
                            ch2[id] = true;
                        }
                    }
                }
            }
            for(int j=0;j<=ttag-1;j++)
            {
                sum[j] = tag[j];
            }
            biao = ttag;
        }
        memset(ch,false,sizeof(ch));
        memset(sum,0,sizeof(sum));
        int ans = dfs(n+1,Top);
        printf("%d %d\n",Top,ans);
    }
    return 0;
}
int binary_search(int l,int r,int val)
{
    int pos;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(f[mid]>=val)
        {
            pos = mid;
            r = mid-1;
        }else
        {
            l = mid+1;
        }
    }
    return pos;
}
void addeage1(int x,int y)
{
    a[Top1].id = y;
    a[Top1].next = c[x];
    c[x] = Top1++;
}
void addeage2(int x,int y)
{
    b[Top2].id = y;
    b[Top2].next = d[x];
    d[x] = Top2++;
}
int dfs(int x,int pt)
{
    if(pt==0)
    {
        return 1;
    }
    if(ch[x])
    {
        return sum[x];
    }
    int res = 0;
    for(int i=d[x];i!=-1;i=b[i].next)
    {
        res+=dfs(b[i].id,pt-1);
    }
    ch[x] = true;
    sum[x] = res;
    return res;
}