hdu 3635 Dragon Balls (并查集)

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2373    Accepted Submission(s): 926

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

 

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  3172 3234 3639 3633 3638 

 

 

//656MS    364K    1191 B    G++
/*

    题意：
        给出n个城市和m个操作，初始每个城市有一颗龙珠，进行若干操作后
     输出某颗龙珠所在城市，及所在城市的龙珠数和某龙珠此时移动步数
     
    并查集：    
        开始时想得挺复杂的，觉得在更新完城市龙珠数时要把原先城市的龙珠数
    清零，可能因为数据问题，这里不清零也不会WA，所以难点就在更新移动步数上，
    在查的时候更新。 

*/
#include<stdio.h>
#include<string.h>
#define N 10005
int set[N],num[N],mov[N];
int find(int x)
{
    if(x!=set[x]){
        int t=set[x];
        set[x]=find(set[x]);
        mov[x]+=mov[t]; //更新值  
    }
    return set[x];
}
void merge(int a,int b)
{
    int x=find(a);
    int y=find(b);
    set[x]=y;
    num[y]+=num[x]; //数量直接加上上一次 
    mov[x]++; //此时移动次数加一 
}
int main(void)
{
    int t,n,m,a,b,k=1;
    char op;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            set[i]=i;  //并查集 
            num[i]=1;  //记录数量 
            mov[i]=0;  //记录步数 
        }
        printf("Case %d:\n",k++);
        for(int i=0;i<m;i++){
            getchar();
            scanf("%c",&op);
            if(op=='T'){
                scanf("%d%d",&a,&b);
                merge(a,b);
            }else{
                scanf("%d",&a);
                b=find(a);
                printf("%d %d %d\n",b,num[b],mov[a]+mov[b]);
            } 
        }
    }
    return 0;
}
/*

6
3 3
T 1 2
T 3 2
Q 2
3 4
T 1 2
Q 1
T 1 3
Q 1
3 4
T 1 2
T 2 1
Q 1
Q 2

*/