【hdu3635】Dragon Balls —— 并查集

题目：

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K
(Java/Others) Total Submission(s): 5393 Accepted Submission(s):
2036

Problem Description Five hundred years later, the number of dragon
balls will increase unexpectedly, so it’s too difficult for Monkey
King(WuKong) to gather all of the dragon balls together.

His country has N cities and there are exactly N dragon balls in the
world. At first, for the ith dragon ball, the sacred dragon will puts
it in the ith city. Through long years, some cities’ dragon ball(s)
would be transported to other cities. To save physical strength WuKong
plans to take Flying Nimbus Cloud, a magical flying cloud to gather
dragon balls. Every time WuKong will collect the information of one
dragon ball, he will ask you the information of that ball. You must
tell him which city the ball is located and how many dragon balls are
there in that city, you also need to tell him how many times the ball
has been transported so far.

Input The first line of the input is a single positive integer T(0 < T
<= 100). For each case, the first line contains two integers: N and Q
(2 < N <= 10000 , 2 < Q <= 10000). Each of the following Q lines
contains either a fact or a question as the follow format: T A B :
All the dragon balls which are in the same city with A have been
transported to the city the Bth ball in. You can assume that the two
cities are different. Q A : WuKong want to know X (the id of the
city Ath ball is in), Y (the count of balls in Xth city) and Z (the
tranporting times of the Ath ball). (1 <= A, B <= N)

Output For each test case, output the test case number formated as
sample output. Then for each query, output a line with three integers
X Y Z saparated by a blank space.

Sample Input 2 3 3 T 1 2 T 3 2 Q 2 3 4 T 1 2 Q 1 T 1 3 Q 1

Sample Output Case 1: 2 3 0 Case 2: 2 2 1 3 3 2

Author possessor WC

Source 2010 ACM-ICPC Multi-University Training Contest（19）——Host by
HDU

描述：

x组用例，每组用例有n个城市和q组指令，初始时每个城市有一个龙珠，指令T A B表示把A城市中的龙珠转移到B城市中，C A表示一个查询，要输出原先在A城市的龙珠现在在的城市编号，该城市中现在的龙珠总数，以及原先在A城市的龙珠被转移了多少次。

题解：

对于查询的前两部分是并查集的裸题，对每一个节点记录他的龙珠数，初始化为1，合并时将子节点的数目加到父节点上再把子节点清零。而对于转移次数的记录，如果每次都在转移时遍历子节点分别次数加1显然会超时，所以这里借鉴了线段树的lazy思想，仅在根节点上time++，当通过子节点查询父节点时，在递归过程中将父节点记录的转移次数加在子节点上。

代码：

#include 
#include 
#include 
using namespace std;
const int maxn = 1e5 + 5;
int fa[maxn];
struct ball
{
    int index,times;
}b[maxn];
void init(int n)
{
    for(int i = 1;i <= n;i++)
    {
        fa[i] = i;
        b[i].times = 0;
        b[i].index = 1;
    }   
}

int find(int x)
{
    if(x == fa[x])return x;
    int temp = fa[x];
    fa[x] = find(fa[x]);
    b[x].times += b[temp].times;
    return fa[x];
}

void unite(int x,int y)
{
    int xx = find(x),yy = find(y);
    if(xx != yy)
    {
        fa[xx] = yy;
        b[xx].times++;
        b[yy].index += b[xx].index;
        b[xx].index = 0;
    } 

}

int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    for(int i = 1;i <= t;i++)
    {
        printf("Case %d:\n",i);
        int n,q;
        scanf("%d%d",&n,&q);
        init(n);
        while(q--)
        {
            char op;
            getchar();
            scanf("%c",&op);
            if(op == 'T')
            {
                int a,b;
                scanf("%d%d",&a,&b);
                unite(a,b);
            }
            else
            {
                int a;
                scanf("%d",&a); 
                int temp = find(a);            
                printf("%d %d %d\n",temp,b[temp].index,b[a].times);
            }
        }
    }
}