poj2761

题意：给出每只狗的pretty value，然后多次询问，每次输出区间[i,j](狗站成一排，从第i只到第j只)的第k小的值是多少。区间之间有交叉，但是没有完全包含。

分析：先把所有区间排序，然后从左至右把每个区间用线段树维护离散化后的pretty value，即线段树的区间的意义是pretty value。每次删除在上一个区间中且不在当前区间中的节点，插入在当前区间中且不在上一个区间中的节点，使得线段树中的节点恰好为该区间内的所有节点。然后查询第k个就容易了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
usingnamespace std;

#define maxn 100005
#define maxm 50005

struct Interval
{
    int s, e, k, pos, ans;
}interval[maxm];

struct Node
{
    Node *pleft, *pright;
    int l, r;
    int num;
}tree[maxn *3];

int dog[maxn], n, m, pretty[maxn], opdog[maxn], ncount;

booloperator< (const Interval &a, const Interval &b)
{
    if (a.s != b.s)
        return a.s < b.s;
    return a.e < b.e;
}

bool cmp(const Interval &a, const Interval &b)
{
    return a.pos < b.pos;
}

int binarysearch(int a)
{
    int l =0;
    int r = n -1;
    int mid = (l + r) /2;
    while (l < r)
    {
        if (pretty[mid] < a)
            l = mid +1;
        else
            r = mid;
        mid = (l + r) /2;
    }
    return l;
}

void buildtree(Node *proot, int l, int r)
{
    proot->l = l;
    proot->r = r;
    proot->num =0;
    if (l == r)
        return;
    int mid = (l + r) /2;
    ncount++;
    proot->pleft = tree + ncount;
    ncount++;
    proot->pright = tree + ncount;
    buildtree(proot->pleft, l, mid);
    buildtree(proot->pright, mid +1, r);
}

void ins(Node *proot, int a, int d)
{
    proot->num += d;
    if (proot->l == proot->r)
        return;
    int mid = (proot->l + proot->r) /2;
    if (a <= mid)
        ins(proot->pleft, a, d);
    else
        ins(proot->pright, a, d);
}

int query(Node *proot, int k)
{
    if (proot->l == proot->r)
        return pretty[proot->l];
    if (proot->pleft->num >= k)
        return query(proot->pleft, k);
    else
        return query(proot->pright, k - proot->pleft->num);
}

void input()
{
    scanf("%d%d", &n, &m);
    for (int i =0; i < n; i++)
    {
        scanf("%d", &dog[i]);
        pretty[i] = dog[i];
    }
    for (int i =0; i < m ; i++)
    {
        scanf("%d%d%d", &interval[i].s, &interval[i].e, &interval[i].k);
        interval[i].s--;
        interval[i].e--;
        interval[i].pos = i;
    }
}

void work()
{
    for (int i = interval[0].s; i <= interval[0].e; i++)
        ins(tree, dog[i], 1);
    interval[0].ans = query(tree, interval[0].k);
    for (int i =1; i < m; i++)
    {
        for (int j = interval[i -1].s; j <= min(interval[i -1].e, interval[i].s -1); j++)
            ins(tree, dog[j], -1);
        for (int j = max(interval[i -1].e +1, interval[i].s); j <= interval[i].e; j++)
            ins(tree, dog[j], 1);
        interval[i].ans = query(tree, interval[i].k);
    }
}

int main()
{
    //freopen("D:\\t.txt", "r", stdin);
    input();
    sort(interval, interval + m);
    sort(pretty, pretty + n);
    for (int i =0; i < n; i++)
    {
        dog[i] = binarysearch(dog[i]);
        opdog[dog[i]] = i;
    }
    ncount =0;
    buildtree(tree, 0, n -1);
    work();
    sort(interval, interval + m, cmp);
    for (int i =0; i < m; i++)
        printf("%d\n", interval[i].ans);
    return0;
}