学步园

题目链接：hdu 5107 K-short Problem

题目大意：有N个点，M次询问，每次询问点X，Y，K，表示在点集合{(x,y)|x≤X,y≤Y}中高度第K小的值是多少，没有的

话输出-1。

解题思路：线段树，每个节点维护10个高度（因为K最大为10），将询问和点按照x，y的大小排序，从左向右，从下向

上，每次询问就查询[0,idx(y)]即可。注意如果询问和点的位置相同，要先插入点。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 60005;

struct node {
    int n, high[15];
    node() { n = 0; }

    void insert(int h) {
        if (n < 10)
            high[n++] = h;
        else if (h < high[n-1])
            high[n-1] = h;
        sort(high, high + n);
    }
};

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2];
node nd[maxn << 2];

inline node merge (node a, node b) {
    node ret;
    int p = 0, q = 0;
    while ((p < a.n || q < b.n) && ret.n <= 10) {
        if (q >= b.n || (p < a.n && a.high[p] < b.high[q]) )
            ret.high[ret.n++] = a.high[p++];
        else
            ret.high[ret.n++] = b.high[q++];
    }
    return ret;
}

void build(int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;

    if (l == r) {
        nd[u].n = 0;
        return ;
    }
    int mid = (l + r) >> 1;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    nd[u] = merge(nd[lson(u)], nd[rson(u)]);
}

void insert (int u, int x, int h) {
    if (x == lc[u] && rc[u] == x) {
        nd[u].insert(h);
        return;
    }

    int mid = (lc[u] + rc[u]) >> 1;
    if (x <= mid)
        insert(lson(u), x, h);
    else
        insert(rson(u), x, h);
    nd[u] = merge(nd[lson(u)], nd[rson(u)]);
}

node query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return nd[u];

    node ret;
    int mid = (lc[u] + rc[u]) >> 1;

    if (l <= mid)
        ret = merge(ret, query(lson(u), l, r));
    if (r > mid)
        ret = merge(ret, query(rson(u), l, r));
    return ret;
}

struct state {
    int x, y, h, id;
}cmd[maxn + 5];

vector<int> pos, tmp;
int N, M, T, ans[30005];

void init () {
    T = N + M;
    pos.clear();
    tmp.clear();
    int x, y, h;

    for (int i = 0; i < N; i++) {
        scanf("%d%d%d", &x, &y, &h);
        tmp.push_back(y);
        cmd[i] = (state) {x, y, h, 0};
    }
    for (int i = 0; i < M; i++) {
        scanf("%d%d%d", &x, &y, &h);
        tmp.push_back(y);
        cmd[i+N] = (state) {x, y, h, i+1};
    }

    sort(tmp.begin(), tmp.end());
    pos.push_back(tmp[0]);
    for (int i = 1; i < tmp.size(); i++) {
        if (tmp[i] != tmp[i-1])
            pos.push_back(tmp[i]);
    }
    build(1, 0, pos.size()-1);
}

inline int find(int x) {
    return lower_bound(pos.begin(), pos.end(), x) - pos.begin();
}

inline bool cmp (const state& a, const state& b) {
    if (a.x != b.x)
        return a.x < b.x;
    if (a.y != b.y)
        return a.y < b.y;
    return a.id < b.id;
}

void solve () {
    sort(cmd, cmd + T, cmp);
    for (int i = 0; i < T; i++) {
        if (cmd[i].id) {
            node tmp = query(1, 0, find(cmd[i].y));
            int kth = cmd[i].h;
            ans[cmd[i].id] = (tmp.n >= kth ? tmp.high[kth-1] : -1);
        } else
            insert(1, find(cmd[i].y), cmd[i].h);
    }
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        init();
        solve();
        for (int i = 1; i <= M; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}