[luogu3377][左偏树(可并堆)]

题目链接

思路

左偏树的模板题，参考左偏树学习笔记
对于这道题我是用一个并查集维护出了哪些点是在同一棵树上，也可以直接log的往上跳寻找根节点

代码

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
inline char nc()
{
    static char buf[1<<14],*p1=buf,*p2=buf;
    return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,1<<14,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
    char c=nc();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=nc();}
    return x*f;
}
int a[N],Pop[N],Col[N];
int n,m;
int ch[N][2],dist[N],fa[N];
int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void uni(int x,int y) {
    x = find(x),y = find(y);
    if(rand() & 1) fa[x] = y;
    else fa[y] = x;
}
int Merge(int x,int y) {
    if(x == 0 || y == 0) return x == 0 ? y : x;
    if(a[x] > a[y] || (a[x] == a[y] && x > y)) swap(x,y);
    ch[x][1] = Merge(ch[x][1],y);
    if(dist[ch[x][1]] > dist[ch[x][0]]) swap(ch[x][1],ch[x][0]);
    dist[x] = dist[ch[x][1]] + 1;
    return x;
}
inline void work1() {
    int x = read(),y = read();
    if(Pop[x] || Pop[y]) return;
    int k1 = Col[find(x)],k2 = Col[find(y)];
    if(k1 == k2) return;
    uni(x,y);
    int z = Merge(k1,k2);
    Col[find(x)] = z;
}
inline void work2() {
    int x = read();
    if(Pop[x]) {
        puts("-1");
        return;
    }
    int k = find(x);
    int y = Col[k];
    printf("%d
",a[y]);
    Pop[y] = 1;
    int now = Merge(ch[y][0],ch[y][1]);
    Col[k] = now;
    ch[y][0] = ch[y][1] = 0;
}
int main() {
    n = read(), m = read();
    dist[0] = -1;
    for(int i = 1;i <= n; ++i) {
        a[i] = read();
        fa[i] = i;
        Col[i] = i;
    }
    while(m--) {
        int bz = read();
        if(bz == 1)  work1();
        if(bz == 2) work2();
    }
    return 0;
}

一言

目标不一定要说出来，去实现它就好。 ——屈屈，