HDU3564：Another LIS(线段树单点+LIS)

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

题意：给出1~n的插入顺序，要求每次插入之后的LIS

这道题还真是花费了不少时间，理解别人代码也花了很久时间，一直在纠结着如何去算LIS，前面还是很简单的，裸的线段树数空格问题，后面LIS的求法看别人的代码确实挺妙，这是用的将每个数的位置存起来，因为数值只有1~n，在这里确实很适合，只要记录了位置，再一次LIS即可。这个思想很巧妙，减一大家自己在演草纸上模拟下，大概思想就是，按照1~n循环下去，这个序列必定是递增的，再比较位置，因为数值递增，所以在保证LIS足够长的情况下，位置小的取小者，最后可以得到最长的值

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,s[1000000],dp[1000000],ans[1000000],len;//ans[i]代表i在位置ans[i];struct node{    int l,r,n;} a[1000000];void init(int l,int r,int i){    a[i].l = l;    a[i].r = r;    if(l == r)    {        a[i].n = 1;        return ;    }    int mid = (l+r)>>1;    init(l,mid,i*2);    init(mid+1,r,i*2+1);    a[i].n = a[i*2].n+a[i*2+1].n;}void insert(int i,int x,int m){    if(a[i].l == a[i].r)    {        ans[m] = a[i].l;        a[i].n=0;        return;    }    a[i].n--;    if(x<=a[2*i].n)        insert(2*i,x,m);    else        insert(2*i+1,x-a[2*i].n,m);}int bin(int k){    int l = 1,r = len;    while(l<=r)    {        int mid = (l+r)>>1;        if(k>dp[mid])            l = mid+1;        else            r = mid-1;    }    return l;}int main(){    int t,i,j,cas = 1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i = 1; i<=n; i++)        {            scanf("%d",&s[i]);            dp[i] = 0;        }        init(1,n,1);        printf("Case #%d:\n",cas++);        for(i = n; i>0; i--)//典型的数空位线段树            insert(1,s[i]+1,i);        len = 0;        for(i = 1; i<=n; i++)//LIS()        {            int k = bin(ans[i]);            len = max(len,k);            dp[k] = ans[i];            printf("%d\n",len);        }        printf("\n");    }    return 0;}