AcWing839. 模擬堆

題解

#include <iostream>

using namespace std;

const int N = 1e5+10;

//hp:堆中下標k,集合中下標為hp[k]
//ph:集合中下標k,堆中下標ph[k]
int p[N], h[N], hp[N], ph[N], cnt, sz;

void heap_swap(int a, int b)
{
    swap(ph[hp[a]], ph[hp[b]]);
    swap(hp[a], hp[b]);
    swap(h[a], h[b]);
}

void down(int u)
{
    int t = u;
    if(u * 2 <= sz && h[u * 2] < h[t]) t = u * 2;
    if(u * 2 + 1 <= sz && h[u * 2 + 1] < h[t]) t = u * 2 + 1;
    if(u != t)
    {
        heap_swap(t, u);
        down(t);
    }
}

void up(int u)
{
    while(u / 2 && h[u / 2] > h[u])
    {
        heap_swap(u, u/2);
        u /= 2;
    }
}

int main()
{
    int n, k, x;
    string c;
    scanf("%d",&n);
    while(n --)
    {
        cin >> c;
        if(c == "I")
        {
            scanf("%d", &k);
            h[++sz] = k;
            p[++cnt] = k;
            hp[sz] = cnt;
            ph[cnt] = sz;
            up(sz);
        }
        else if(c == "PM") printf("%d\n",h[1]);
        else if(c == "DM")
        {
            heap_swap(1, sz);
            sz -- ;
            down(1);
        }
        else if(c == "D")
        {
            scanf("%d",&k);
            k = ph[k];
            heap_swap(k, sz);
            sz -- ;
            down(k);
            up(k);
        }
        else{
            scanf("%d%d", &k, &x);
            h[ph[k]] = x;
            down(ph[k]);
            up(ph[k]);
        }
    }
    return 0;
}