poj 2201 Cartesian Tree

http://poj.org/problem?id=2201

　　正如题目所说，这是笛卡尔树的构造，输出每个结点的父节点，左右子结点的标号。十分简单的题，轻松1y！

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

using namespace std;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<piii> vpiii;

const int maxn = 50001;

struct Node{
    int key, fix;
    int id;
    int child[2], parent;
    Node(int _key = 0, int _fix = 0, int _id = 0){
        child[0] = child[1] = parent = 0;
        key = _key;
        fix = _fix;
        id = _id;
    }
}node[maxn];

struct Cartesian{
    int root, back;
    Cartesian(){
        root = back = 0;
    }

    void pushBack(int x){
        int p = back;

        while (p && node[p].fix > node[x].fix){
            p = node[p].parent;
        }
        if (p){
            if (node[p].child[false]) node[node[p].child[false]].parent = x;
            node[x].child[true] = node[p].child[false];
            node[x].parent = p;
            node[p].child[false] = x;
        }
        else{
            if (root) node[root].parent = x;
            node[x].child[true] = root;
            root = x;
        }
        back = x;
    }

    void print(int T){
        if (!T) return ;
        printf("%d : left %d right %d parent %d key %d id %d\n", T, node[T].child[1], node[T].child[0], node[T].parent, node[T].key, node[T].id);
        print(node[T].child[1]);
        print(node[T].child[0]);
    }
}cart;

vpiii buf;
int mark[maxn];

void deal(int n){
    sort(buf.begin(), buf.end());
    cart = Cartesian();

    int size = buf.size();

    for (int i = 0; i < size; i++){
        node[i + 1] = Node(buf[i].first.first, buf[i].first.second, buf[i].second);
        cart.pushBack(i + 1);
//        cart.print(cart.root);
//        puts("!!!end");
    }
    for (int i = 1; i <= n; i++){
        mark[node[i].id] = i;
    }


    for (int i = 1; i <= n; i++){
        printf("%d %d %d\n", node[node[mark[i]].parent].id, node[node[mark[i]].child[1]].id, node[node[mark[i]].child[0]].id);
    }
}

int main(){
    int n;
    int key, fix;

    while (~scanf("%d", &n)){
        buf.clear();
        puts("YES");
        for (int i = 0; i < n; i++){
            scanf("%d%d", &key, &fix);
            buf.push_back(make_pair(make_pair(key, fix), i + 1));
        }
        deal(n);
    }

    return 0;
}

——written by Lyon