探讨使用并查集和左偏树解决猴子冲突问题，包括输入输出样例及算法实现细节。
Total Submit: 1511 Accepted Submit: 390
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
解题报告
从题目的描述可以看出，需要判断两只猴子是否属于同一集合并且将不同集合的猴子合并为一个集合，这我们可以想到并查集。同时我们需要求出一个集合中力量最大的猴子，将力量最大猴子的力量减半，然后将两个集合进行合并并重新选择力量最大的猴子。开始我用的是二叉堆，结果就是TLE。原因在于在一个堆中逐个插入元素并进行调整的时间复杂度为O(nlogn)。因此需要一种支持堆之间快速合并操作的可合并堆。这我们可以选用两种：左偏树以及二项堆。由于二项堆过于复杂，在此我选用的是左偏树。每个并查集合的根节点记录集合对应左偏树的根节点。具体步骤如下：
1．初始化并查集和左偏树；
2 ．猴子x与y之间有conflict
2.1 利用并查集判断x与y是否在同一个集合中，如果是，输出-1；
2.2 通过两只猴子所在并查集集合的根节点获取猴子所在的左偏树的根节点hx和hy。hx和hy的力量减半；
2.3 合并hx和hy。由于hx和hy的力量减半，因此hx和hy可能已不满足左偏树的性质（节点的键值大于等于左右子 树 根节点的键值），不能直接合并。我最先采用的方法是写个下降调整的函数，先将两边调整为左偏树，然后合 并，但是老TLE，最后不得不换成另外一种方法：先将hx和hy从各自所在的左偏树中去掉，然后将hx的左右子树合并形成左偏树nx，hy的左右子树合并形成左偏树mx，接着将mx合并到nx；hx和hy从所在左偏树去掉后，形成两棵只有一个节点的左偏树，将hy合并到hx，最后hx合并到nx就形成了一颗新的左偏树。虽然这样比较繁琐，但是可以看到这若干步合并操作总的时间复杂度只有log(n)+log(m)，n和m分别是原来两颗左偏树的节点个数。第一种方法的时间复杂度就多出了下降调整的开销log(n)+log(m)，因此时间开销是第二种方法的两倍。
2.4 合并两只猴子所在的并查集集合，并重新记录集合中力量最大的猴子。
代码：
#include
<stdio.h>#include
<string.h>#define
MAXN 100000typedef
struct{
}HeapNode;
int val;int dist;int left;int right;static
HeapNode H[MAXN+1];static
int parent[MAXN+1];static
int heapRef[MAXN+1];int
{
t = x;
{
k = parent[t];
parent[t] = root_x;
t = k;
}
}
Find_Set(int x)int root_x;int t,k;for(root_x = x; parent[root_x]>=0; root_x = parent[root_x]);while(t!=root_x)return root_x;int
{
{
parent[x] += parent[y];
parent[y] = x;
p = x;
}
Union(int x, int y)int p;if(parent[x] < parent[y])else
{
parent[y] += parent[x];
parent[x] = y;
p = y;
}
}
return p;//ri<-rj
static
{
{
ri = rj;
rj = temp;
}
H[ri].right = merge(H[ri].right,rj);
{
H[ri].left = H[ri].right;
H[ri].right = temp;
}
H[ri].dist = H[right].dist + 1;
}
int merge(int ri, int rj)if(ri == 0)return rj;if(rj == 0)return ri;if(H[ri].val < H[rj].val)int temp = ri;int left = H[ri].left;int right = H[ri].right;if(H[left].dist < H[right].dist)int temp = H[ri].left;return ri;int
{
H[0].dist = -1;
H[0].val = -1;
{
{
scanf(
H[i].left = H[i].right = 0;
H[i].dist = 0;
parent[i] = -1;
heapRef[i] = i;
}
scanf(
{
scanf(
rx = Find_Set(x);
ry = Find_Set(y);
{
printf(
}
hx = heapRef[rx];
hy = heapRef[ry];
H[hx].val/=2;
H[hy].val/=2;
hxl = H[hx].left;
hxr = H[hx].right;
hyl = H[hy].left;
hyr = H[hy].right;
nx = merge(hxl,hxr);
mx = merge(hyl,hyr);
nx = merge(nx,mx);
H[hx].left = H[hx].right = 0;
H[hx].dist = 0;
H[hy].left = H[hy].right = 0;
H[hy].dist = 0;
hx = merge(hx,hy);
nx = merge(nx,hx);
j = Union(rx,ry);
heapRef[j] = nx;
printf(
}
}
}
欢迎各位指正！
网上别人给出了二项堆的做法，不过比较复杂，我把它贴在下面：
扫一扫
116
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
