HDOJ 1512 几乎模板的左偏树

题目大意：有n个猴子。每个猴子有一个力量值，力量值越大表示这个猴子打架越厉害。如果2个猴子不认识，他们就会找他们认识的猴子中力量最大的出来单挑，单挑不论输赢，单挑的2个猴子力量值减半，这2拨猴子就都认识了，不打不相识嘛。现在给m组询问，如果2只猴子相互认识，输出-1，否则他们各自找自己认识的最牛叉的猴子单挑，求挑完后这拨猴子力量最大值。

左偏大根加并查

代码都懒的贴了...

#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
const int maxn = 100010;
struct node {
        int l,r,dis,val,par;
} heap[maxn];
int N, M;
int find ( int &x ) {
    return heap[x].par == x ? x : heap[x].par = find ( heap[x].par );
}
int merge(int rt1,int rt2)  
{  
    if (rt1==0) return rt2;  
    if (rt2==0) return rt1;  
    if ( heap[rt2].val>heap[rt1].val )swap(rt1,rt2);  
    heap[rt1].r = merge(heap[rt1].r,rt2);
    heap[heap[rt1].r].par = rt1;
    if ( heap[ heap[rt1].l ].dis < heap[ heap[rt2].r ].dis ) 
         swap ( heap[rt1].l, heap[rt1].r ); 
    else heap[rt1].dis = heap[ heap[rt1].r ].dis + 1;
    return rt1;  
}  
int push ( int x, int y ) {
    return merge ( x, y );       
}
int pop ( int &x ) {
    int l = heap[x].l; 
    int r = heap[x].r; 
    heap[l].par = l;
    heap[r].par = r;
    heap[x].l = heap[x].r = heap[x].dis = 0;   
    return merge ( l, r );  
}
bool scan_d(int &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}
int main() {
    while ( scan_d ( N ) ) {
         for ( int i = 1; i <= N; ++ i ) {
              scan_d ( heap[i].val );
              heap[i].l = heap[i].r = heap[i].dis = 0;
              heap[i].par = i;    
         }
         scan_d ( M );
         int a, b, x, y;
         while ( M -- ) {
                scan_d (a); scan_d (b);
                x = find ( a );
                y = find ( b ); 
                if ( x == y ) {
                    puts ( "-1" );     
                } else {
                    heap[x].val /= 2;
                    int px = push ( pop ( x ), x );  
                    heap[y].val /= 2;
                    int py = push ( pop ( y ), y );  
                    
                    printf ( "%d
", heap[ merge ( px, py ) ].val );      
                }    
         } 
    }
    return 0;
}