Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2909    Accepted Submission(s): 1125

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

23 3T 1 2T 3 2Q 23 4T 1 2Q 1T 1 3Q 1

Sample Output

Case 1:2 3 0Case 2:2 2 13 3 2

：：挺不错的一道题，并查集，个人觉得维护某个球移动次数最难想到怎么去维护。

对于一个点的移动次数只要在合并的时候把该点移动次数加上其父亲的移动次数就好了。（想想，只有移动次数为0的球才能作为一个集合的“根”）；

下面的代码思想是一样的，只是后一种用了点小技巧，省空间

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int N  = 10010; 7 int _, cas=1, n, m, fa[N], num[N], shift[N]; 8  9 void init()10 {11     for(int i=1; i<=n; i++) fa[i]=i, num[i]=1, shift[i]=0;12 }13 14 int find(int x)15 {16     if(x==fa[x]) return x;17     int p = fa[x];18     fa[x] = find(fa[x]);19     shift[x] += shift[p];20     return fa[x];21 }22 23 void move_to(int u, int v)24 {25     u = find(u) , v =find(v);26     if(u==v) return ;27     fa[u] = v;28     num[v] += num[u];29     shift[u]++;30 }31 32 void solve()33 {34     scanf("%d%d", &n, &m);35     init();36     char s[3];37     int u, v;38     printf("Case %d:\n", cas++);39     while(m--)40     {41         scanf("%s%d", s, &u);42         if(s[0]==‘T‘){43             scanf("%d", &v);44             move_to(u, v);45         }46         else{47             v =find(u);48             printf("%d %d %d\n", v, num[v], shift[u]);49         }50     }51 }52 53 int main()54 {55 //    freopen("in.txt", "r", stdin);56     cin>>_;57     while(_--) solve();58     return 0;59 }

View Code

view code#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N  = 10010;int _, cas=1, n, m, fa[N], shift[N];int find(int x){    if(fa[x]<0) return x;    int p = fa[x];    fa[x] = find(fa[x]);    shift[x] += shift[p];    return fa[x];}void move_to(int u, int v){    u = find(u) , v =find(v);    if(u==v) return ;    fa[v] += fa[u];    fa[u] = v;    shift[u]++;}void solve(){    scanf("%d%d", &n, &m);    for(int i=1; i<=n; i++) fa[i]=-1, shift[i]=0;    char s[3];    int u, v;    printf("Case %d:\n", cas++);    while(m--)    {        scanf("%s%d", s, &u);        if(s[0]==‘T‘){            scanf("%d", &v);            move_to(u, v);        }        else{            v =find(u);            printf("%d %d %d\n", v, -fa[v], shift[u]);        }    }}int main(){//    freopen("in.txt", "r", stdin);    cin>>_;    while(_--) solve();    return 0;}