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
题意:
分别在题目给的位置插入1-N个数字,求每一次插入后的最长递增子序列。
这道题对与初学线段树的简直可以说是噩梦,无奈只好搜题解。
发现竟是如此的巧妙。
建立一个线段树,表示区间内的空位数。
然后倒着遍历数组所需要插入的位置,倒着插入的好处是不用挪动其他元素了,因为后面的元素的位置已经确定好了。 线段树是区间求和。
然后通过一个数组no记录下1-N数字的位置。
接下来是对no求最长连续递增子序列。
这里用的是二分查找法, 时间复杂度为N*lonN。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100005;
const int INF=0x3f3f3f3f;
int t;
int tree[maxn<<2];
int a[maxn];
int no[maxn];
int dp[maxn];
int n;
void pushup (int re)
{
tree[re]=tree[re<<1]+tree[re<<1|1];
}
void build (int l,int r,int re)
{
if(l==r)
{
tree[re]=1;
return ;
}
int mid=(l+r)>>1;
build (l,mid,re<<1);
build (mid+1,r,re<<1|1);
pushup (re);
}
//找到插入位置并更新线段树
void update (int data,int l,int r,int re,int zz)
{
if(l==r)
{
tree[re]--;
no[zz]=l;
return;
}
int mid=(l+r)>>1;
if(data<=tree[re<<1])
update (data,l,mid,re<<1,zz);
else
update (data-tree[re<<1],mid+1,r,re<<1|1,zz);
pushup (re);
}
//二分查找
int finds (int a,int len)
{
int l=1,r=len;
while (l<=r)
{
int mid=(l+r)>>1;
if(dp[mid]<=a)
l=mid+1;
else if(dp[mid]==a)
return mid;
else
r=mid-1;
}
return l;
}
int main()
{
scanf("%d",&t);
for (int i=1;i<=t;i++)
{
scanf("%d",&n);
build (1,n,1);
printf("Case #%d:\n",i);
for (int j=1;j<=n;j++)
{
scanf("%d",&a[j]);
}
for (int j=n;j>=1;j--)
{
update (a[j]+1,1,n,1,j);
}
//求最长递增子序列
int len=1;
dp[1]=no[1];
printf("1\n");
for (int j=2;j<=n;j++)
{
if(dp[len]<no[j])
{
dp[++len]=no[j];
}
else
{
int pos=finds(no[j],len);
dp[pos]=no[j];
}
printf("%d\n",len);
}
printf("\n");
}
return 0;
}