【POJ 2104】K-th Number 题意&题解&代码（c++）

K-th Number

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Time Limit: 20000MS Memory Limit: 65536K
Case Time Limit: 2000MS

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Description

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You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1…n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: “What would be the k-th number in a[i…j] segment, if this segment was sorted?”
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2…5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

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Input

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The first line of the input file contains n — the size of the array, and m — the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 109 by their absolute values — the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

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Output

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For each question output the answer to it — the k-th number in sorted a[i…j] segment.

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Sample Input

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7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

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Sample Output

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5
6
3

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Hint

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This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

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中文题意

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非常简洁：给出一数组a，然后要求查询从 a[i] 到 a[j] 的第k大的数。

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题解：需要运用到主席树这种坑爹算法。可以到网上了解一下。
我的代码中也有相应的注释。

推荐网站站：
http://www.cnblogs.com/AbandonZHANG/archive/2012/09/27/2706000.html

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代码：

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#include
#include
#include
#include
using namespace std;
int tot=0,root[100005];
//tot记录节点的编号，root[i]表示第i棵线段树根节点的编号
map<int,int>hash;
map<int,int>mp;
//了解主席树之后可以知道主席树所占用的内存是很大的因此需要hash后建树
//这样可以尽可能的减少空间的使用
struct edge{
    int l;int r;//表示线段树上这个节点所代表区间l~r
    int ls;int rs;//表示这个节点左儿子和右儿子的编号
    int val;//这个节点所代表的区间中有多少个数
}tree[2000005];
int build_tree(int l,int r)//建最原始的空树
{
    tot++;
    int k=tot;
    tree[k].l=l;
    tree[k].r=r;
    tree[k].val=0;
    if (l==r) return k;
    int mid=(l+r)/2;
    tree[k].ls=build_tree(l,mid);
    tree[k].rs=build_tree(mid+1,r);
    return k;
}
int add_tree(int id,int x,int v)//建新树
{
    tot++;
    int k=tot;

    tree[k]=tree[id];
    //每次只改变 从需要更改的节点 到 根结点 的一条链
    //因此 另一个儿子 可以直接等于 前一个线段树 的 儿子
    //这样可以减少许多的空间
    tree[k].val+=v;

    if ( tree[id].r==x && tree[id].l==x )
    return k;

    int mid=(tree[id].l+tree[id].r)/2;

    if (x<=mid)//判断需要改哪条链
    tree[k].ls=add_tree(tree[id].ls,x,v);
    else
    tree[k].rs=add_tree(tree[id].rs,x,v);

    return k;
}
int query_tree(int ne,int no,int w)
{
    int tmp=tree[tree[no].ls].val-tree[tree[ne].ls].val;
    //建好n棵线段树之后可以发现第i棵线段树与第i+1棵线段树的区别是
    //是否在线段树中加了第i+1个点
    //因此要 求区间l和r中所出现的数字 可以用tree[r]与tree[l-1]做差
    //找到做差后第k个数字即可
    if (tree[no].l==tree[no].r)
    return tree[no].l;

    if (tmp>=w)
    return query_tree(tree[ne].ls,tree[no].ls,w);
    else
    return query_tree(tree[ne].rs,tree[no].rs,w-tmp);
}
int n,m,p[200005],s[200005];
int main()
{
    scanf("%d",&n);
    scanf("%d",&m);

    for (int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        s[i]=p[i];
    }
    sort(p+1,p+1+n);
    int num=0;
    for (int i=1;i<=n;i++)
    {
        if (hash[p[i]]==0)
        {
            num++;
            hash[p[i]]=num;
            mp[num]=p[i];
        }   
    }
    root[0]=build_tree(1,num);

    for (int i=1;i<=n;i++)
    {
        int cha=hash[s[i]];
        root[i]=add_tree(root[i-1],cha,1);
    }

    for (int i=1;i<=m;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        int ans=query_tree(root[a-1],root[b],c);
        printf("%d\n",mp[ans]);
    }
}