POJ 1988 Cube Stacking(并查集)两种解法
作者：互联网
地址：http://poj.org/problem?id=1988
Description
Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.
Write a program that can verify the results of the game.
Input
* Line 1: A single integer, P* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.
Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.
Output
Print the output from each of the count operations in the same order as the input file.Sample Input
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
题意：n个操作。M x y 表示将x所在的那堆箱子放在y那堆箱子头上。C x 询问x下面有多少箱子。
解析一：以deep[x]记录x上面有多少箱子，堆顶为根，以all[x]记录x所在堆有多少个箱子。初始化为deep[]=0,all[]=1,pr[i]=i。find()函数的更新过程（包含路径压缩）
int find(int x)
{
if(pr[x]==x)return x;
int mid=pr[x];
pr[x]=find(pr[x]);//路径压缩
deep[x]+=deep[mid]; //下图记录此步
return pr[x];
}
分析过程：
接下来是join():
void join(int a,int b)
{
int fa=find(a),fb=find(b);
if(fa!=fb)
{
pr[fb]=fa;//以堆顶为根，a放在了b的头上
deep[fb]+=all[fa];//下图分析此过程
all[fa]+=all[fb];//下图分析此过程
}
return ;
}
分析过程：
计算当前点下面有多少个箱子，就是：all[find(x)]-deep[x]-1
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=3e5+50;
int n;
int pr[maxn],deep[maxn],all[maxn];
void init()
{
for(int i=1;i<=n;i++)
{
pr[i]=i;
deep[i]=0;
all[i]=1;
}
}
int find(int x)
{
if(pr[x]==x)return x;
int mid=pr[x];
pr[x]=find(pr[x]);
deep[x]+=deep[mid];
return pr[x];
}
void join(int a,int b)
{
int fa=find(a),fb=find(b);
if(fa!=fb)
{
pr[fb]=fa;
deep[fb]+=all[fa];
all[fa]+=all[fb];
}
return ;
}
int main()
{
while(~scanf("%d",&n))
{
init();
char ch;
int mid=n;
while(mid--)
{
scanf(" %c",&ch);
if(ch=='M')
{
int a,b;
scanf("%d%d",&a,&b);
join(a,b);
}
else
{
int x;
scanf("%d",&x);
cout<<all[find(x)]-deep[x]-1<<endl;
}
}
}
}
解析二：以deep[x]记录x下面有多少箱子，堆底为根。其他与上面基本相同。find()相同。不同的是join()谁接谁改了一下：
void join(int a,int b)
{
int fa=find(a),fb=find(b);
if(fa!=fb)
{
pr[fa]=fb;//堆底为根。
deep[fa]+=all[fb];//与上面不同
all[fb]+=all[fa];//与上面不同
}
return ;
}
询问x，那么find(x)一下，防止之前没更新，直接输出deep[x]即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=3e5+50;
int n;
int pr[maxn],deep[maxn],all[maxn];
void init()
{
for(int i=1;i<=n;i++)
{
pr[i]=i;
deep[i]=0;
all[i]=1;
}
}
int find(int x)
{
if(pr[x]==x)return x;
int mid=pr[x];
pr[x]=find(pr[x]);
deep[x]+=deep[mid];
return pr[x];
}
void join(int a,int b)
{
int fa=find(a),fb=find(b);
if(fa!=fb)
{
pr[fa]=fb;
deep[fa]+=all[fb];
all[fb]+=all[fa];
}
return ;
}
int main()
{
while(~scanf("%d",&n))
{
init();
char ch;
int mid=n;
while(mid--)
{
scanf(" %c",&ch);
if(ch=='M')
{
int a,b;
scanf("%d%d",&a,&b);
join(a,b);
}
else
{
int x;
scanf("%d",&x);
find(x);
cout<<deep[x]<<endl;
}
}
}
}
标签：pr,fa,Stacking,查集,deep,int,1988,include,find 来源： https://www.cnblogs.com/liyexin/p/12627267.html