Double Queue 【POJ - 3481】【SET】

题目链接

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  知道这道题可以用splay tree去做，但是先想到的肯定会是set的解法，然后再去考虑splay的做法，那么就先放上set的做法了。

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#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
struct node
{
    int K, P;
    node(int a=0, int b=0):K(a), P(b) {}
    friend bool operator == (node e1, node e2) { return e1.K == e2.K && e1.P == e2.P; }
    friend bool operator < (node e1, node e2) { return e1.P > e2.P; }
};
set<node> st;
set<node>::iterator it;
int main()
{
    int op, k, p;
    while(scanf("%d", &op) && op)
    {
        if(op == 1)
        {
            scanf("%d%d", &k, &p);
            st.insert(node(k, p));
        }
        else if(op == 2)
        {
            if(!st.size()) { printf("0\n"); continue; }
            it = st.begin();
            printf("%d\n", it->K);
            st.erase(it);
        }
        else
        {
            if(!st.size()) { printf("0\n"); continue; }
            it = --st.end();
            printf("%d\n", it->K);
            st.erase(it);
        }
    }
    return 0;
}