Another LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

There is a sequence firstly empty. We begin to add number from 1 to N to the sequence, and every time we just add a single number to the sequence at a specific position. Now, we want to know length of the LIS (Longest Increasing Subsequence)
after every time‘s add.

Input

An integer T (T <= 10), indicating there are T test cases.

For every test case, an integer N (1 <= N <= 100000) comes first, then there are N numbers, the k-th number Xk means that we add number k at position Xk (0 <= Xk <= k-1).See hint for more details.

Output

For the k-th test case, first output "Case #k:" in a separate line, then followed N lines indicating the answer. Output a blank line after every test case.

Sample Input

1
3
0 0 2

Sample Output

Case #1:
1
1
2

Hint

In the sample, we add three numbers to the sequence, and form three sequences.
a. 1
b. 2 1
c. 2 1 3


 

Author

standy

Source

2010 ACM-ICPC Multi-University Training
Contest（13）——Host by UESTC

解题：先求出这个序列+排序+LIS。

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 100100
typedef struct nnn
{
    int id,i;
}point;
int tree[3*N];
point oder[N];
void bulide1(int l,int r,int k)
{
    tree[k]=r-l+1;
    if(l==r)return ;
    bulide1(l,(l+r)/2,k*2);
    bulide1((l+r)/2+1,r,k*2+1);
}
void set_tree1(int l,int r,int k,int id,int i)
{
    int m=(l+r)/2;
    tree[k]--;
    if(l==r)
    {
        oder[l].id=l; oder[l].i=i; return ;
    }
    if(tree[k*2]>=id)set_tree1(l,m,k*2,id,i);
    else set_tree1(m+1,r,k*2+1,id-tree[k*2],i);
}

bool cmp(point a,point b){return a.i<b.i;}
void bulide2(int l,int r,int k)
{
    tree[k]=0;
    if(l==r)return ;
    bulide2(l,(l+r)/2,k*2);
    bulide2((l+r)/2+1,r,k*2+1);
}
void set_tree2(int l,int r,int k,int id,int lis)
{
    int m=(l+r)/2;
    if(l==r)
    {
       if(tree[k]<lis) tree[k]=lis; return ;
    }
    if(id<=m)set_tree2(l,m,k*2,id,lis);
    else set_tree2(m+1,r,k*2+1,id,lis);

    if(tree[k*2]>tree[k*2+1])
        tree[k]=tree[k*2];
    else tree[k]=tree[k*2+1];
}
int query2(int l,int r,int k,int id)
{
    int m=(l+r)/2;
    if(l==r)return 0;
    int max=0;
    if(m>=id)
       max=query2(l,m,k*2,id);
    else
    {
        max=query2(m+1,r,k*2+1,id);
        if(max<tree[k*2]) max=tree[k*2];//左边的数一定比当前数小，因为排了个序从小到大排
    }
    return max;
}
int main()
{
    int a[N+5],LIS[N+5],m,n;
    scanf("%d",&m);
    for(int j=1;j<=m;j++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]); a[i]++;
        }

        bulide1(1,n,1);
        for(int i=n;i>0;i--)
            set_tree1(1,n,1,a[i],i);

        sort(oder+1,oder+n+1,cmp);
        bulide2(1,n,1);
        for(int i=1;i<=n;i++)
        {
            int lis=1+query2(1,n,1,oder[i].id);
            set_tree2(1,n,1,oder[i].id,lis);
            LIS[i]=tree[1];
        }
        printf("Case #%d:\n",j);
        for(int i=1;i<=n;i++)
            printf("%d\n",LIS[i]);

        printf("\n");
    }
}