hdu1512 & zoj2334Monkey King(左偏树 + 并查集)

Monkey King

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2543    Accepted Submission(s): 1064

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
   
   
   
   

 

Author

JIANG, Yanyan

 

Source

ZOJ 3rd Anniversary Contest

 

Recommend

linle

题目大意：有n个猴子，一开始每个猴子只认识自己。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。

题目分析：首先很明显这题涉及到集合并的操作，要用到并查集。其次要找到某一拨猴子中力量最大值，找最大值最快的应该是堆。2拨猴子要快速合并而又不失堆的特性，想来想去左偏树比较合适。

关于左偏树，百度写的很详细。大致说两句。

首先左偏树是一个堆，具体什么堆看题目要求，比如这题就要求一个大堆。

其次左偏树顾名思义还有左偏的性质。这里的左偏不是说树的左子树就一定比右子树大。左偏树每个节点除了堆的关键字key外，还有一个关键字dis，距离，这是左偏树的特色。左偏树定义这样一类节点：如果一个节点的左右子树有空树，称之为外点，树根到一个外点的最近距离为dis。比如一颗叶子节点的dis为0。左偏树就是根据dis值来调整姿势的。左偏树保证左偏的性质：节点的左子树的dis值要不小于右子树的dis值。也就是说左偏树从根节点一路向右能最快访问到外点。

所以左偏树其实就是一颗有左偏性质的堆有序二叉树。

左偏树并不是一颗平衡树。因为左偏树的主要功能是快速访问最值节点以及修改后的快速恢复。所以左偏树放弃了平衡 的性质。相反，一颗平衡的左偏树反而会影响左偏树的性能。

左偏树详细内容可以看这篇左偏树的特点及其应用，讲的十分详细。

关于这题，用一个并查集维护猴子的集合，左偏树用来快速合并，并且快速保持大堆性质。并查集的并操作其实可以在左偏树合并的时候进行。

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100005;
int m,n;
int set[N];
struct node
{
    int l,r,dis,key;
}tree[N];

int find(int a)
{
    int root = a;
    while(root != set[root])
        root = set[root];
    int parent = set[a];
    while(parent != root)//路径压缩
    {
        set[a] = root;
        a = parent;
        parent = set[a];
    }
    return root;
}

int merge(int a,int b)
{
    if(!a)
        return b;
    if(!b)
        return a;
    if(tree[a].key < tree[b].key)//大堆
        swap(a,b);
    tree[a].r = merge(tree[a].r,b);
    set[tree[a].r] = a;//并查
    if(tree[tree[a].l].dis < tree[tree[a].r].dis)
        swap(tree[a].l,tree[a].r);
    if(tree[a].r)
        tree[a].dis = tree[tree[a].r].dis + 1;
    else
        tree[a].dis = 0;
    return a;
}

int pop(int a)
{
    int l = tree[a].l;
    int r = tree[a].r;
    set[l] = l;//因为要暂时删掉根，所以左右子树先作为根
    set[r] = r;
    tree[a].l = tree[a].r = tree[a].dis = 0;
    return merge(l,r);
}

int nextint()
{
    char c;
    int ret = 0;
    while((c = getchar()) > '9' || c < '0')
        ;
    ret = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        ret = ret * 10 + c - '0';
    return ret;
}

void print(int a)
{
    if(!a)
        return;
    print(a/10);
    putchar(a%10 + '0');
}

int main()
{
    int a,b,i;
    while(~scanf("%d",&n))
    {
        for(i = 1;i <= n;i ++)
        {
            //scanf("%d",&tree[i].key);
            tree[i].key = nextint();
            set[i] = i;
            tree[i].l = tree[i].r = tree[i].dis = 0;
        }
       // scanf("%d",&m);
        m = nextint();
        while(m --)
        {
            //scanf("%d%d",&a,&b);
            a = nextint();
            b = nextint();
            int ra = find(a);
            int rb = find(b);
            //printf("%d  %d\n",ra,rb);
            if(ra == rb)
                printf("-1\n");
            else
            {
                int rra = pop(ra);//ra左右子树合并
                tree[ra].key /= 2;
                ra = merge(rra,ra);//重新插入ra 找到合适的位置
                int rrb = pop(rb);
                tree[rb].key /= 2;
                rb = merge(rrb,rb);
                print(tree[merge(ra,rb)].key);
                putchar(10);
                //printf("%d\n",tree[merge(ra,rb)].key);
            }
        }
    }
    return 0;
}
//703MS	2244K  无优化
//250MS	2184K  输入优化
//203MS	2184K  输入输出优化