洛谷P3377 【模板】左偏树（可并堆）

30分代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
inline int read(){
    int x = 0,tmp = 1;char ch = getchar();
    while( ch < '0' || ch > '9' ) {if ( ch == '-' ) tmp = -1; ch = getchar();}
    while( ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar(); }
    return x * tmp;
}
inline void write( int k ) {
    if ( k < 0 ) { putchar( '-' ); k = -k;}
    if ( k > 9 ) write( k / 10 );
    putchar( k % 10 + '0' );
}
struct tree{
    int l, r, v, dis, f;
} heap[110000];
bool hash[110000];
inline int find( int x ) {
    return heap[x].f == x ? x : find( heap[x].f );
}
inline int merge( int a, int b ) {
    if ( heap[a].v > heap[b].v || a > b && heap[a].v == heap[b].v) swap( a, b );
    if( a == 0 ) return b;
    heap[a].r = merge( heap[a].r, b );
    heap[heap[a].r].f = a;
    if( heap[heap[a].l].dis < heap[heap[a].r].dis ) swap( heap[a].l, heap[a].r );
    //if( heap[a].r == 0 ) heap[a].dis = 0; else
    heap[a].dis = heap[heap[a].r].dis + 1;
    return a;
}
inline int pop( int x ){
    int l = heap[x].l, r = heap[x].r;
    heap[l].f = l;
    heap[r].f = r;
    heap[x].l = heap[x].r = heap[x].dis = 0;
    return merge( l ,r );
}
int main() {
    int N = read(), M = read();
    for( int i=1; i<=N ; ++i) {
        heap[i].v = read();
        heap[i].l = heap[i].r = heap[i].dis = 0;
        heap[i].f = i;
    }
    for( int i=1; i<=M; ++i) {
        int flag = read();
        if( flag == 1 ) {
            int x = read(), y = read();
            int findx = find ( x ), findy = find ( y );
            if ( findx == findy ) continue;
            merge( findx, findy );
        } else if( flag == 2 ) {
            int x = read();
            int findx = find( x );
            if ( !(hash[x] || hash[findx]) ) {
                write( heap[findx].v );
                putchar( '\n' );
                pop( findx );
                hash[findx] = 1;        
            } else printf( "-1\n" );
        }
    }

    return 0;
}

struct Point {
    int to, next;
} edge[210000];
int head[110000], inde[110000], idx = 0, f[110000];
queue<int> Q;
inline void ade( int u, int v ) {
    inde[v] ++;
    edge[++ idx].to = v;
    edge[idx].next = head[u];
    head[u] = idx;
}
int main(){
    memset( f, 0, sizeof( f ) );
    memset( head, -1, sizeof( head) );
    int N = read(), M = read();
    for( int i = 1 ; i <= M ; ++ i ) {
        int x = read(), y = read();
        ade( x, y );
    }
    for( int i = 1 ; i <= N ; ++ i )
        if( !inde[i] ) {
            Q.push( i );
            f[i] = 1;
        }
    while( !Q.empty() ) {
        int u = Q.front();
        Q.pop();
        for( int i = head[u] ; i != -1 ; i = edge[i].next ) {
            int v = edge[i].to;
            inde[v] --;
            f[v] = max( f[v], f[u] + 1 );
            if( !inde[v] ) Q.push( v );
         }
    } 
    for( int i = 1 ; i <= N ; ++ i ) {
        write( f[i] );
        putchar( '\n' );
    }
    
    return 0;
}