Dragon bils
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 bils will increase unexpectedly, so it's too difficult for Monkey King (Wukong) to gather all of the dragon bils together.
His country has n cities and there are exactly n dragon bils 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) wocould be transported to other cities. to save physical strength Wukong plans to take flying nimbus cloud, a magical flying cloud to gather dragon bils.
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 bils 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 bils 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 Ballin 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 3 T 1 2 T 3 2q 23 4 T 1 2q 1 T 1 3q 1
Sample output
Case 3 0 case 2 13 3 2
: This is a good question, and I check the set. I personally think it is the most difficult to maintain a ball movement.
For the number of moves of a vertex, you only need to add the number of moves of the vertex to the number of moves of its father when merging. (Think About It. Only balls with 0 moves can serve as the "root" of a set ");
The following code has the same idea, but the latter uses some tips to save space.
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;}