ZOJ 2334 Monkey King

并查集+左偏树。。。。。合并的时候用左偏树，合并结束后吧父结点全部定成树的根节点，保证任意两个猴子都可以通过Find找到最厉害的猴子

                      Monkey King

Time Limit: 10000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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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 their 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

Source

ZOJ 3rd Anniversary Contest

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=100100;

int father[maxn];

struct leafistreenode
{
    int d,v,l,r;
}heap[maxn];

int merge(int x,int y)
{
    if(!x) return y; if(!y) return x;
    if(heap[x].v<heap[y].v) swap(x,y); ///big root heap
    heap[x].r=merge(heap[x].r,y);
    int l=heap[x].l,r=heap[x].r;
    if(heap[l].d<heap[r].d) swap(heap[x].l,heap[x].r);
    if(!heap[x].r) heap[x].d=0;
    else heap[x].d=heap[r].d+1;
    return x;
}

int N,M;

int Find(int x)
{
    if(father[x]!=x) return father[x]=Find(father[x]);
}

int solve(int a,int b)
{
    int x=Find(a),y=Find(b);
    if(x==y) return -1;

    int p,xx,yy,temp;

    heap[x].v/=2;
    temp=merge(heap[x].l,heap[x].r);
    heap[x].d=heap[x].l=heap[x].r=0;
    xx=merge(temp,x);

    heap[y].v/=2;
    temp=merge(heap[y].l,heap[y].r);
    heap[y].d=heap[y].l=heap[y].r=0;
    yy=merge(temp,y);

    p=merge(xx,yy);

    father[xx]=father[yy]=p=father[x]=father[y]=father[a]=father[b]=p;

    return heap[p].v;
}

int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        memset(heap,0,sizeof(heap));
        for(int i=0;i<=N+10;i++) father[i]=i;
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&heap[i].v);
        }
        scanf("%d",&M);
        while(M--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d
",solve(a,b));
        }
    }
    return 0;
}