笛卡尔树模版

按第一关键字排序，然后按第二关键字构建笛卡尔树即可。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
template <typename A>
inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 60000;
int n, x, y;
vector< pair<pair<int, int>, int> > v;
struct Node{
    int i, val;
    Node *fa, *lf, *rf;
    Node(){}
    Node(int i, int val, Node *fa, Node *lf, Node *rf): i(i), val(val), fa(fa), lf(fa), rf(rf){}
}*null, *root;

bool cmp(Node *a, Node *b){ return a->i < b->i; }

Node *buildTree(){

    stack<Node *> st;
    Node *last = null;
    for(int i = 0; i < v.size(); i ++){
        Node *cur = new Node(v[i].second, v[i].first.second, null, null, null);
        while(!st.empty()){
            if(st.top()->val < cur->val){
                Node *up = st.top();
                if(up->rf != null) cur->lf = up->rf, up->rf->fa = cur;
                up->rf = cur, cur->fa = up;
                break;
            }
            last = st.top(), st.pop();
        }
        if(st.empty() && last) cur->lf = last, last->fa = cur;
        st.push(cur);
    }
    Node *root = null;
    while(!st.empty()) root = st.top(), st.pop();
    return root;
}
vector<Node *> ans;

void dfs(Node *cur){
    if(cur == null) return;
    ans.pb(cur);
    dfs(cur->lf), dfs(cur->rf);
}

int main(){

    n = read();
    for(int i = 0; i < n; i ++){
        x = read(), y = read();
        v.pb(make_pair(make_pair(x, y), i + 1));
    }
    sort(v.begin(), v.end());
    null = new Node(0, 0, NULL, NULL, NULL);
    root = buildTree();
    dfs(root);
    sort(ans.begin(), ans.end(), cmp);
    puts("YES");
    for(int i = 0; i < ans.size(); i ++){
        printf("%d %d %d\n", ans[i]->fa->i, ans[i]->lf->i, ans[i]->rf->i);
    }
    return 0;
}

标签：Node,2201,Cartesian,cur,int,st,rf,POJ,ans
来源： https://www.cnblogs.com/onionQAQ/p/11345058.html