Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4517    Accepted Submission(s): 1719

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

主要是并查集，对并查集的理解，然后加点对移动次数的记录，只需每次加上自己父节点的移动个数即可！

即：移动次数是把自己的加上父亲结点的就是移动总数了

具体看代码：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=10010;
int F[MAXN];
int num[MAXN];
int cnm[MAXN];
int n;
void init()
{
    for(int i=1;i<=n;i++)
    {
        F[i]=i;//父节点
        num[i]=1;// 记录点数
        cnm[i]=0;//移动次数
    }
}
int find(int x)
{
	int t;
    if(F[x]==x)
	return x;
     t=F[x];
    F[x]=find(F[x]);
    cnm[x]+=cnm[t];
    return F[x];
}
void bing(int a,int b)
{
    int t1=find(a);
    int t2=find(b);
    if(t1!=t2)
    {
        F[t1]=t2;
        num[t2]+=num[t1];//点数加
        cnm[t1]=1;//将移动的点加入计算
    }
}

int main()
{
    int m;
    int T;
    char str[10];
    int a,b;
    int iCase=0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d%d",&n,&m);
        init();
        printf("Case %d:\n",iCase);
        while(m--)
        {
            scanf("%s",&str);
            if(str[0]=='T')
            {
                scanf("%d%d",&a,&b);
                bing(a,b);
            }
            else
            {
                scanf("%d",&a);
                int t=find(a);
                printf("%d %d %d\n",t,num[t],cnm[a]);
            }
        }
    }
    return 0;
}

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