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 = ;
 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==) return rt2;
     if (rt2==) 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 + ;
     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 = ;
     return merge ( l, r );
 }
 bool scan_d(int &num) {
     char in;bool IsN=false;
     in=getchar();
     if(in==EOF) return false;
     while(in!='-'&&(in<''||in>'')) in=getchar();
     if(in=='-'){ IsN=true;num=;}
     else num=in-'';
     while(in=getchar(),in>=''&&in<=''){
             num*=,num+=in-'';
     }
     if(IsN) num=-num;
     return true;
 }
 int main() {
     while ( scan_d ( N ) ) {
          for ( int i = ; i <= N; ++ i ) {
               scan_d ( heap[i].val );
               heap[i].l = heap[i].r = heap[i].dis = ;
               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 /= ;
                     int px = push ( pop ( x ), x );
                     heap[y].val /= ;
                     int py = push ( pop ( y ), y );  

                     printf ( "%d\n", heap[ merge ( px, py ) ].val );
                 }
          }
     }
     return ;
 }