hdu3564(二分+线段树)

Another LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 936    Accepted Submission(s): 328

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

130 0 2

 

Sample Output

Case #1:112
Hint
In the sample, we add three numbers to the sequence, and form three sequences.a. 1b. 2 1c. 2 1 3
 

 

Author

standy

 

Source

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

 

Recommend

zhouzeyong

        本题给定一个数的序列，其中a[i]表示在第a[i]个位置插入i。输入的位置是没有顺序的。因此不能用一般的方法模拟插入，时间复杂度O(n^2)。可以用线段树。设置线段树的节点信息包含一个num表示该区间[left,right]还有的空位个数。最后一个数插入的位置a[i]肯定就是正确的位置；然后倒数第二个考虑已插入的数……于是可得最终的序列。但题目的LIS是动态的，即在每次插入后计算输出。线段树很容易球的每个元素的真实位置。依题意，最终序列da[1-n]存放的是1-n的某个排序。da[i]表示元素i在第da[i]个位置。我们从i=1开始求LIS，求得的时对应下表的LIS，但与元素的LIS等价。于是我们要求的是da[]的LIS，可以用二分的方法求解，时间复杂度为O(N*log(N))。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;//************************************************************//算法1的时间复杂度为O(N*log(N))const int MAXN=100000+100;int data[MAXN];//存放原始数据int MaxV[MAXN];//MaxV[i]存放长度为i的严格递增子序列的最大值的最小值int ans[MAXN];int da[MAXN];struct node {int left,right,num;//num此处灵活处理}tree[MAXN*4];//1.建立以left,right为左右边界,将数组da中元素存储在首地址从1开始的线段树tree的叶节点上void Build( int id,int left,int right){    tree[id].left=left;    tree[id].right=right;tree[id].num=right-left+1;//此处灵活处理    if(left==right)return ;    else    {int mid =(left+right)>>1;                Build(id<<1,left,mid);        Build((id<<1)|1,mid+1,right);   }}void Updata(int id,int pos,int val){    tree[id].num--;    if(tree[id].left==tree[id].right)    {da[val]=tree[id].left;//val表示第几次操作，da[】存储的位置        return ;    }if(tree[id<<1].num>=pos)         Updata(id<<1,pos,val);    else Updata((id<<1)|1,pos-tree[id<<1].num,val);}//二分查找返回MaxV中大于等于x的组靠左的下标int BinaryResearch(int x,int len){int mid,low=1,high=len;while(low<=high){mid=(low+high)>>1;if(MaxV[mid]<x)low=mid+1;else high=mid-1;}return low;}//返回原序列中严格递增子序列的最长长度void LIS(int n){int i,len=0;for(i=1;i<=n;i++){if(i==0||da[i]>MaxV[len])//比长度为len的子序列最大值大,直接加进末尾MaxV[++len]=da[i];else {int pos=BinaryResearch(da[i],len);MaxV[pos]=da[i];}ans[i]=len;}}//===============================================int main(){int cas,i,n,tag=1;cin>>cas;while(cas--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&data[i]);data[i]++;}Build(1,1,n);for(i=n;i>=1;i--){Updata(1,data[i],i);}LIS(n);printf("Case #%d:\n",tag++);for(i=1;i<=n;i++)printf("%d  ",da[i]);printf("*******\n");for(i=1;i<=n;i++)printf("%d\n",ans[i]);printf("\n");}system("pause");return 0;}