HDU3564 --- Another LIS （线段树维护最值问题）

Another LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1244    Accepted Submission(s): 431

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

 

题意：一个空序列，第k次在xk处 插入 k，并输出 序列当前的LIS。

首先 倒序 用线段树来确定每个元素 最终的位置。然后 求n次LIS，当然要用线段树维护了，，或者二分等等也可以。 nlogn

 #include <cstdio>
 #include <string>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int maxn = 1e5+;
 int seg[maxn<<],a[maxn];            // seg表示该区间 位置剩余量
 void build (int l,int r,int pos)
 {
     if (l == r)
     {
         seg[pos] = ;
         return;
     }
     int mid = (l + r) >> ;
     build(l,mid,pos<<);
     build(mid+,r,pos<<|);
     seg[pos] = seg[pos<<] + seg[pos<<|];
 }
 int ans[maxn];                            //ans[i]表示 元素i的位置为ans[i]
 void Insert(int l,int r,int pos,int x,int k)
 {
     if (l == r)
     {
         ans[k] = l;
         seg[pos] = ;
         return ;
     }
     int mid = (l + r) >> ;
     if (seg[pos<<] >= x)                            //左孩子的区间 位置剩余量大于x时，递归进入左孩子，否则右孩子同时 x-seg[pos<<1]
         Insert(l,mid,pos<<,x,k);
     else
         Insert(mid+,r,pos<<|,x - seg[pos<<],k);
     seg[pos] = seg[pos<<] + seg[pos<<|];
 }
 void update(int l,int r,int pos,int x,int v)
 {
     if (l == r)
     {
         seg[pos] = v;
         return;
     }
     int mid = (l + r) >> ;
     if (x <= mid)
         update(l,mid,pos<<,x,v);
     else
         update(mid+,r,pos<<|,x,v);
     seg[pos] = max(seg[pos<<],seg[pos<<|]);
 }
 int query(int l,int r,int pos,int ua,int ub)
 {
     if (ua <= l && ub >= r)
     {
         return seg[pos];
     }
     int mid = (l + r) >> ;
     int t1 = ,t2 = ;
     if (ua <= mid)
         t1 = query(l,mid,pos<<,ua,ub);
     if (ub > mid)
         t2 = query(mid+,r,pos<<|,ua,ub);
     return max(t1,t2);
 }

 int main(void)
 {
     #ifndef ONLINE_JUDGE
         freopen("in.txt","r",stdin);
     #endif
     int t,cas = ;
     scanf ("%d",&t);
     while (t--)
     {
         int n;
         scanf ("%d",&n);
         for (int i = ; i <= n; i++)
             scanf ("%d",a+i);
         build (,n,);
         for (int i = n; i >= ; i--)
             Insert(,n,,a[i] + ,i);               //倒序 确定每个元素 最终的位置保存在ans里
         memset(seg,,sizeof(seg));
         printf("Case #%d:\n",cas++);
         for (int i = ; i <= n; i++)
         {
             int t1 = query(,n,,,n);
             int t2 = query(,n,,,ans[i]);
             update(,n,,ans[i],t2+);
             if (t2 < t1)                        //输出当前区间的LIS
                 printf("%d\n",t1);
             else
                 printf("%d\n",+t2);
         }
         printf("\n");
     }
     return ;
 }