HDU3564 Another LIS 线段树

Another LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1419    Accepted Submission(s): 521

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 

http://acm.split.hdu.edu.cn/showproblem.php?pid=3564

 题意 把1~N个数依次插入相应的位置 有个地方恶心 如样例把1放到0位置 再把2放到0位置 1就必须相应的往后面挪

显然后面插入的大数优先级高  用线段树数空格的方法从后往前插入

最后把得到一个排好的序列   如果用普通的最长上升子序列的方法应该会超时 n*n/2

看了一下大神的 很巧妙吧    http://blog.csdn.net/libin56842/article/details/13095801

每次放的都是一个大于以前所有数的数者这样LiS必然大于等于以前 就看它在什么位置了

如果在len后面必然增加一

#include<stdio.h>#include<algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MAX 100010int tree[MAX*4],a[MAX],ans[MAX],dp[MAX];int len;//ans[i] 记录是的i的位置ans[i]void pushup(int rt){    tree[rt]=tree[rt<<1]+tree[rt<<1|1];}void build(int l,int r,int rt){    if(l==r)    {        tree[rt]=1;        return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int p,int x,int l,int r,int rt){    if(l==r)    {        ans[x]=l;        tree[rt]=0;        return ;    }    tree[rt]--;  //每次消耗一    int m=(l+r)>>1;    if(p<=tree[rt<<1]) update(p,x,lson);//左边的位置够就左边    else    {        p=p-tree[rt<<1];       //不够就到右边相应的位置去找        update(p,x,rson);    }}int fan(int k){    int l=1,r=len,mid;    while(l<=r)    {    mid=(l+r)>>1;        if(k>dp[mid])            l=mid+1;        else r=mid-1;    }    return l;}int main(){    int t,i,n,k,tt;    scanf("%d",&t);      tt=1;    while(t--)    {        scanf("%d",&n);        for(i=1; i<=n; i++)        {            scanf("%d",&a[i]);            dp[i]=0;            a[i]++;        }        build(1,n,1);        for(i=n; i>0; i--)            update(a[i],i,1,n,1);//数空位线段树        len=0;        printf("Case #%d:\n",tt++);        for(i=1; i<=n; i++)        {            k=fan(ans[i]);            len=max(len,k);            dp[k]=ans[i];            printf("%d\n",len);        }        printf("\n");    }    return 0;}