Cube Stacking

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 17959 Accepted: 6208Case Time Limit: 1000MS

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

6M 1 6C 1M 2 4M 2 6C 3C 4

Sample Output

102

代码：

#include<iostream>#include<cstdio>#define Maxn 30010using namespace std;int fa[Maxn],dis[Maxn],height[Maxn];int findset(int x){    if(fa[x]!=x){        int t=fa[x];        fa[x]=findset(fa[x]);        dis[x]+=dis[t];    }    return fa[x];}void unionset(int x,int y){    x=findset(x),y=findset(y);    if(x==y) return;    fa[y]=x;    dis[y]=height[x];    height[x]+=height[y];}int main(){    int p,a,b;    char s;    cin>>p;    for(int i=0;i<Maxn;i++) fa[i]=i,dis[i]=0,height[i]=1;    for(int i=0;i<p;i++){        cin>>s;        if(s=='M') {cin>>a>>b;unionset(a,b);}        else{            cin>>a;            int t=findset(a);            cout<<height[t]-dis[a]-1<<endl;        }    }    return 0;}

注：这题又是一道种类并查集，思想和食物链那道类似，关键这题的突破口是想到每个元素与并查集根的距离，通过每堆总结点数，即可换算出需要的结果，维护距离是在路径压缩时，这样可以做的原因，在我看来是，需要用到这点，这点必被查找必然更新。像这种并查集有带向量运算感，因此叫偏移向量也不为过。这题很明显距离是直接可加的，而食物链那题并不明显，那几个公式都要枚举所用情况，总结得到的，而且还带有模运算，但最终的公式回归到了向量运算。

我一直在想这是巧合吗？还是必然……其实我到现在还没完全想明白，但我更偏向巧合，因为食物链那题只有三种关系，并且必须设同类关系为0，否则就没有矢量性，而且巧合的是三种不同类生物构成一个矢量三角形……如果想要推广到n种关系，那么就不止要满足矢量n边形了，最重要的是必须满足加法性质，而且加完一圈可以绕回来，甚至从n点中任取m（m>=2）点，都需满足上述性质，这样才会有偏移矢量。

以上纯属个人看法，不喜勿喷！欢迎大牛前来指正！