加权并查集——（POJ 1988）Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 25647		Accepted: 8975
Case 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

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

Sample Output

1
0
2

Source

USACO 2004 U S Open

思路：

　　加权并查集

　　题意： 说是有n块砖，编号从1到n,有两种操作，第一是把含有x编号的那一堆砖放到含有编号y的那一堆砖的上面，第二是查询编号为x的砖的下面有多少块砖。用count[x]表示下面有多少块砖。

　　现在需要把两堆砖合并，显然要用上并查集，可是普通的合并之后如何知道x的下面有多少块砖呢，思考合并的过程，对于一堆砖，移动到另一堆砖上时，上面那一堆上每块砖的count[i]应该加上下面一堆砖的数量，这个操作对于上面一堆砖的根来说是简单的，我使用uset[i]表示连通分量，舒适化时所有的uset[i]为-1,负数代表这个节点为根，1代表这个连通分量的节点总数为1,以样例为例，首先将1放到6上面，即将6合并到1所在的连通分量中，合并的过程中我们知道两个信息，第一是当前连通分量6->1的节点数量为2,6距离1的距离为1,同理，将2放到4上面，这个连通分量节点个数为2,,4到2的距离为1,最后，我们将包含6的这个连通分量合并到包含2的这个连通分量中，此时连通分量数为4，曾经的6->1连通分量的根距离合并后的连通分量的根的距离为2,就是4->2的连通分量的节点数

　　说了半天有什么用处呢，经过上面这个过程，我们知道了每一个节点到它第一次被合并时的那个根节点的距离，6->1的距离为1,1到4的距离为2,2到4的距离为1,这样我们在查询6的下面有多少块砖时，直接用4(连通分量节点数)-（1+2）（6到根节点的距离）-1

上代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 30000+10;
int fa[maxn];
int rank[maxn];
int dist[maxn];

void init() {
    for(int i=0; i<maxn; i++) {
        fa[i]=i;
        rank[i] = 1;
        dist[i] = 0;
    }
}

int find(int x) {
    if(x != fa[x]) {
        int t = fa[x];
        fa[x] = find(fa[x]);
        dist[x] += dist[t];//dis[t]表示t，即x曾经的根节点距离根的距离，x到根的距离为x到t的距离加上t到根的距离
    }
    return fa[x];
}

int main() {
    int n;
    while(scanf("%d",&n)==1) {
        init();
        char op;
        while(n--) {
            cin>>op;
            int a,b;
            if(op=='M') {
                scanf("%d%d",&a,&b);
                int faA = find(a);
                int faB = find(b);
                if(faA != faB) {
                    fa[faB]=faA;
                    dist[faB]=rank[faA];//下面的砖的根距离如今的连通分量的根的距离
                    rank[faA]+=rank[faB];//节点相加
                }
            } else {
                scanf("%d",&a);
                int x = find(a);
                printf("%d
",rank[x]-dist[a]-1);//连通分量节点数-根节点的距离-1 
            }
        }
    }
    return 0;
}

自己选的路，跪着也要走完！！！