HDU 1512 HDOJ 1512 Monkey King ( 左偏树 ) ACM 1512 IN HDU

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋    

 

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1512

题目描述 :

Monkey King



Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 914    Accepted Submission(s): 426





Problem Description

Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know each other. And when it happens, both the two monkeys will invite the strongest friend of them, and duel. Of course, after the duel, the two monkeys and all of there friends knows each other, and the quarrel above will no longer happens between these monkeys even if they have ever conflicted.



Assume that every money has a strongness value, which will be reduced to only half of the original after a duel(that is, 10 will be reduced to 5 and 5 will be reduced to 2).



And we also assume that every monkey knows himself. That is, when he is the strongest one in all of his friends, he himself will go to duel.

 



Input

There are several test cases, and each case consists of two parts.



First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).



Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth.



 



Output

For each of the conflict, output -1 if the two monkeys know each other, otherwise output the strongness value of the strongest monkey in all friends of them after the duel.

 



Sample Input

5

20

16

10

10

4

5

2 3

3 4

3 5

4 5

1 5

 



Sample Output

8

5

5

-1

10

 

 

 

题目分析:

/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_1512
Doc Name  : Monkey King
    
    
题目意思: 



有N只猴子, 每只都有一个力量值. 开始的时候互不认识, 它们之间会发生M次斗争. 每次发生a, b的斗争时, a, b都会从各自的朋友圈里拉出一个最强的, 之后两只猴子打, 打完后这两只猴子的力量值各减半. 并且打完后, 两只猴子的朋友圈的所有人都互相认识(也就是不会再打).



你的任务就是对于每个斗争, 若a, b是朋友, 那么输出-1, 否则输出打完后它们的朋友圈的最强猴子的力量值.



 使用 普通 优先队列的话 估计会超时, 因为数据量很大 100000 ! !, 等下有空试试看. 



对于每一个节点, 定义dis 表示X节点到最右边的空节点的距离的最小值



对于每个节点X, 要求X的左儿子的dis >= 右儿子的dis, 那么容易发现, 对于N个节点的左偏树, 其右儿子最多只有logN个节点.



合并操作就是让复杂度落在右儿子上, 从而达到logN的合并复杂度.



首先对于两个堆, 若其中一个为空, 返回另一个.



否则(这里以大根堆为例), a指向堆顶较大的堆, b指向另一个. 让a的右儿子和b合并, 合并后的子树作为a的右儿子.



接下来, 检查a的两个儿子是否满足dis, 不满足就交换两个儿子.



最后, 更新a的dis.



这样就容易实现堆的其他操作 ( 比如插入, 删除顶等 ).



另外 还需要用到 并查集.    
    
    
*/
//#pragma warning( disable:4789 )
#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 MM = 100010;
struct left {
        int l,r,dis,val,dad;
} heap[MM];



int N, M;



inline int max ( const int &a, const int &b) {
       return a > b ? a : b;
}



inline int find ( int &x ) {
    return heap[x].dad == x ? x : heap[x].dad = find ( heap[x].dad );
}



inline void swap(int &a, int &b) {
     a ^= b ^= a ^= b;
}



inline int merge ( int x, int y ) {
    if ( x == 0 ) return y;
    if ( y == 0 ) return x;
    if ( heap[y].val > heap[x].val ) swap ( x, y );    
    heap[x].r = merge ( heap[x].r, y );
    heap[heap[x].r].dad = x;
    if ( heap[ heap[x].l ].dis < heap[ heap[x].r ].dis ) 
         swap ( heap[x].l, heap[x].r );
    if ( heap[x].r == 0 ) heap[x].dis = 0;
    else heap[x].dis = heap[ heap[x].r ].dis + 1;
    return x;
}



inline int push ( int x, int y ) {
       return merge ( x, y );       
}



inline int pop ( int &x ) {
       int l = heap[x].l; 
       int r = heap[x].r; 
       heap[l].dad = l;
       heap[r].dad = r;
       heap[x].l = heap[x].r = heap[x].dis = 0;   
       return merge ( l, r );  
}



inline 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].dad = 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 xx = push ( pop ( x ), x );  
                    heap[y].val /= 2;
                    int yy = push ( pop ( y ), y );  
                    
                    printf ( "%d\n", heap[ merge ( xx, yy ) ].val );      
                }    
         } 
    }
    return 0;
}