中南OJ1551: Longest Increasing Subsequence Again（分块+离散化线段树）

Description

Give you a numeric sequence. If you can demolish arbitrary amount of numbers, what is the length of the longest increasing sequence, which is made up of consecutive numbers? It sounds like Longest Increasing Subsequence at first sight. So, there is another limitation: the numbers you deleted must be consecutive.

Input

There are several test cases.
For each test case, the first line of input contains the length of sequence N(1≤N≤10^4). The second line contains the elements of sequence——N positive integers not larger than 10^4.

Output

For each the case, output one integer per line, denoting the length of the longest increasing sequence of consecutive numbers, which is achievable by demolishing some(may be zero) consecutive numbers.

Sample Input

7 1 7 3 5 9 4 8 6 2 3 100 4 4 5

Sample Output

4 4

Source

#include<stdio.h>
#include<algorithm>
using namespace std;
const int N = 10005;
struct qj
{
    int l,r;
}kuai[N];
int b[N],tn,n,a[N],k[N],tree[N*4];
void builde(int l,int r,int e)
{
    int m=(l+r)/2;
    tree[e]=0;
    if(l==r)
        return ;
    builde(l,m,e*2);
    builde(m+1,r,e*2+1);
}
void settree(int l,int r,int e,int id,int ans)
{
    int m=(l+r)/2;
    if(tree[e]<ans)
        tree[e]=ans;
    if(l==r)
    {
         return ;
    }
    if(id<=m)
        settree(l,m,e*2,id,ans);
    else
        settree(m+1,r,e*2+1,id,ans);
}
int findans(int l,int r,int e,int id)
{
    int m=(l+r)/2,ans=0;
    if(l==r)
        return 0;
    if(id<=m)
    {
        ans=findans(l,m,e*2,id);
        if(ans<tree[e*2+1])
            ans=tree[e*2+1];
        return ans;
    }
    else
        return findans(m+1,r,e*2+1,id);
}
int cmp(int aa,int bb)
{
    return aa<bb;
}
void cut()
{
    tn=1;
    sort(b+1,b+1+n,cmp);
    for(int i=2;i<=n;i++)
        if(b[i]!=b[tn])
            b[++tn]=b[i];
}
int two(int aa)
{
    int l,r,m;
    l=1;r=tn;
    while(l<=r)
    {
        m=(l+r)/2;
        if(b[m]==aa)
            return m;
        if(b[m]>aa)
            r=m-1;
        else l=m+1;
    }
    return m;
}
int main()
{
    while(scanf("%d",&n)>0)
    {
        if(n==0)
        {
            printf("0\n");continue;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        cut();
        builde(1,tn,1);

        int m=1;
        kuai[m].l=kuai[m].r=1;
        for(int i=2;i<=n;i++)
         if(a[kuai[m].r]>=a[i])
        {
            m++; kuai[m].l=kuai[m].r=i;
        }
        else
            kuai[m].r++;

        for(int i=1;i<=m;i++)
        {
            for(int j=kuai[i].r;j>=kuai[i].l;j--)
                k[j]=kuai[i].r-j+1;
        }
        int ans=0;
        
        for(int i=m;i>0;i--)
        {
            int aaa,id;
            for(int j=kuai[i].r;j>=kuai[i].l;j--)
            {
                id=two(a[j]);
                aaa=j-kuai[i].l+1;
                aaa+=findans(1,tn,1,id);
                if(ans<aaa)
                ans=aaa;
            }

            for(int j=kuai[i].r;j>=kuai[i].l;j--)
            {
                id=two(a[j]);
                settree(1,tn,1,id,k[j]);
            }
        }
        printf("%d\n",ans);
    }
}

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原文地址：http://blog.csdn.net/u010372095/article/details/44707473