hdu 3635 Dragon Balls(並查集)

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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

：：挺不錯的一道題，並查集，個人覺得維護某個球移動次數最難想到怎麼去維護。

對於一個點的移動次數只要在合并的時候把該點移動次數加上其父親的移動次數就好了。（想想，只有移動次數為0的球才能作為一個集合的“根”）；

 

下面的代碼思想是一樣的，只是後一種用了點小技巧，省空間

 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;}