Dragon Balls

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

Total Submission(s): 3245 Accepted Submission(s): 1252

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

题意：有n个城市，开始时每个城市只有一个球，编号1到n。两种操作，一种是（T，a，b），把第a个球所在城市的所有球运送到b号球所在的城市，另一种是（Q，a），查询第a个球运送到了第X个城市，第X个城市有Y个球，第a个球被运输了Z次，输出X Y Z。

思路：并查集，union过程维护一个球总数sum，以及find过程维护运输次数t

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                using namespace std;const int INF=1<<30;const double eps=1e-6;const int N = 10000;int fa[N],sum[N],t[N];int n,m;int find(int x){ if(fa[x]==x) return x; else { int tmp=find(fa[x]); t[x]+=t[fa[x]]; return fa[x]=tmp; }}void run(){ static int cas = 1; printf("Case %d:\n",cas++); char s[5]; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) fa[i]=i,sum[i]=1,t[i]=0; int u,v; while(m--) { scanf("%s",s); if(s[0]=='T') { scanf("%d%d",&u,&v); u=find(u); v=find(v); if(u!=v) { fa[u]=v; sum[v]+=sum[u]; t[u]++; } } else { scanf("%d",&u); v=find(u); printf("%d %d %d\n",v,sum[v],t[u]); } }}int main(){ freopen("case.txt","r",stdin); int _; scanf("%d",&_); while(_--) run(); return 0;}

转载于:https://www.cnblogs.com/someblue/p/3944445.html

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