HDOJ 题目3564 Another LIS（线段树单点更新，LIS）

Another LIS

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

Total Submission(s): 1291 Accepted Submission(s): 451



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

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思路：就是从左插入找空位，从1~n用线段树记录他们的位置，然后再对他们的位置进行LIS就好

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int a[100010];
int node[100010<<2],d[100010],len,dp[100010];
void build(int l,int r,int tr)
{
node=r-l+1;
if(l==r)
return;
int mid=(l+r)>>1;
build(l,mid,tr<<1);
build(mid+1,r,tr<<1|1);
node=node[tr<<1]+node[tr<<1|1];
}
int bin(int x)
{
int l=1,r=len;
while(l<=r)
{
int mid=(l+r)>>1;
if(x>dp[mid])
l=mid+1;
else
r=mid-1;
}
return l;
}
void insert(int pos,int num,int l,int r,int tr)
{
if(l==r)
{
d[num]=l;
node=0;
return;
}
int mid=(l+r)>>1;
node--;
if(pos<=node[tr<<1])
{
insert(pos,num,l,mid,tr<<1);
}
else
insert(pos-node[tr<<1],num,mid+1,r,tr<<1|1);
}
int main()
{
int t,c=0;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int i;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
dp[i]=0;
}
build(1,n,1);
for(i=n;i>0;i--)
{
insert(a[i]+1,i,1,n,1);
}
len=0;
/*for(i=1;i<=n;i++)
{
printf("%d\n",d[i]);
}*/
printf("Case #%d:\n",++c);
for(i=1;i<=n;i++)
{
int k=bin(d[i]);
len=max(len,k);
dp[k]=d[i];
printf("%d\n",len);
}
printf("\n");
}
}