hdu2665 Kth number（主席树模板）

Problem Description
Give you a sequence and ask you the kth big number of a inteval.

Input
The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output
For each test case, output m lines. Each line contains the kth big number.

Sample Input
1
10 1
1 4 2 3 5 6 7 8 9 0
1 3 2

Sample Output
2

分析：
主席树模板
http://blog.csdn.net/metalseed/article/details/8045038
这个blog讲的很好

主席树实际上就是一棵权值线段树
按照权值从小到大新建树
每个元素在树中按照原序列的顺序安置
这就能保证在x之前的且比x小的元素，都有记录
画个图模拟一下，加深印象

tip

询问区间第k小
空间不要开炸
主席树的空间一般是n*40
但是还要视情况而定

这道题n*40就MLE了，啊啊啊啊

hdu爆炸的情况挺严重。。。

这里写代码片#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N=100010;struct node{    int l,r,sum;};node tree[N*20];int num[N],n,m,root[N],top=0;struct nd{    int v,bh;};nd a[N];int cmp(const nd &a,const nd &b){return a.v<b.v;}void insert(int &now,int l,int r,int pm){    top++;    tree[top]=tree[now];    now=top;    tree[now].sum++;    if (l==r)        return;    int mid=(l+r)>>1;    if (pm<=mid) insert(tree[now].l,l,mid,pm);    else insert(tree[now].r,mid+1,r,pm);}int ask(int x,int y,int l,int r,int k){    if (l==r)        return l;   //返回的是在重排后的序列中的位置     int t=tree[tree[y].l].sum-tree[tree[x].l].sum;    int mid=(l+r)>>1;    if (t>=k) return ask(tree[x].l,tree[y].l,l,mid,k);    else return ask(tree[x].r,tree[y].r,mid+1,r,k-t);}int main(){    int T;    scanf("%d",&T);    while (T--)    {        scanf("%d%d",&n,&m);        for (int i=1;i<=n;i++) scanf("%d",&a[i].v),a[i].bh=i;        sort(a+1,a+1+n,cmp);        for (int i=1;i<=n;i++) num[a[i].bh]=i;           //num[i]离散后排名是i的元素在原序列中的排名          //重复元素???        for (int i=1;i<=n;i++)   //按权值插入         {            root[i]=root[i-1];            insert(root[i],1,n,num[i]);        }          for (int i=1;i<=m;i++)        {            int x,y,k;            scanf("%d%d%d",&x,&y,&k);            printf("%d\n",a[ask(root[x-1],root[y],1,n,k].v);  //第k大         }    }    return 0;}