UVA 11987 Almost Union-Find (并查集）

Almost Union-Find

I hope you know the beautiful Union-Find structure. In this problem, you're to implement something similar, but not identical.

The data structure you need to write is also a collection of disjoint sets, supporting 3 operations:

1 p q

Union the sets containing p and q. If p and q are already in the same set, ignore this command.

2 p q

Move p to the set containing q. If p and q are already in the same set, ignore this command

3 p

Return the number of elements and the sum of elements in the set containing p.

Initially, the collection contains n sets: {1}, {2}, {3}, ..., {n}.

Input

There are several test cases. Each test case begins with a line containing two integers n and m (1<=n,m<=100,000), the number of integers, and the number of commands. Each of the next m lines contains a command. For every operation, 1<=p,q<=n. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each type-3 command, output 2 integers: the number of elements and the sum of elements.

Sample Input

5 71 1 22 3 41 3 53 42 4 13 43 3

Output for the Sample Input

3 123 72 8

题意：给定n个城市，m种操作。接下来m行输入m组操作

1为连接城市a，b。2为把a拿出来（即a的所有连接都断掉），再连接a，b。3为输出该点的连接中共有几点和这些点的和。

思路：并查集的使用。。这题多了个删除所以麻烦点。我是用2个数组parent和father来查找父节点。

然后操作1时直接连接没什么好说的。。操作2的时候。利用另一个数组father来把该点连接到另一片图中。

这样最后在输出的时候。只要查找每个点的father的根节点的情况就可以了。。。

#include <stdio.h>#include <string.h>int n, m;int sb;int x, y;int dian;int parent[100005];int father[100005];int sum[100005];int num[100005];int find(int x){    if (parent[x] != x)return parent[x] = find(parent[x]);    elsereturn x;}int main(){    while (scanf("%d%d", &n, &m) != EOF)    {int i;for (i = 1; i <= n ; i ++){    parent[i] = i;    father[i] = i;    num[i] = 1;    sum[i] = i;}for (i = 0; i < m ; i++){scanf("%d", &sb);if (sb == 1){scanf("%d%d", &x, &y);if (father[x] == father[y])continue;int pa = find(father[x]);int pb = find(father[y]);if (pa == pb)continue;num[pb] += num[pa];sum[pb] += sum[pa];parent[pa] = pb;}else if (sb == 2){scanf("%d%d", &x, &y);if (father[x] == father[y])continue;int pa = find(father[x]);int pb = find(father[y]);if (pa == pb)continue;father[x] = father[y];num[pa] --;num[pb] ++;sum[pa] -= x;sum[pb] += x;}else if (sb == 3){scanf("%d", &dian);printf("%d %d\n", num[find(father[dian])], sum[find(father[dian])]);}}    }    return 0;}