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
#include<stdio.h>int parent[101010];int num[101001];//表示移动次数 int rank[100010];//表示该地有多少颗龙珠 int find(int x){if(x==parent[x]) return x;int t=parent[x];parent[x]=find(parent[x]);//压缩路径 ，都指向根节点num[x]+=num[t];//每个球移动的次数等于本身移动的个数加上父节点移动的次数return parent[x]; //int r=x;//while(r!=parent[r])//{//num[r]+=num[parent[r]]; //num需要通过加上父节点来更新 //r=parent[r];//}//int i=x,j;//路径压缩 //while(i!=r)//{//j=parent[i];//parent[i]=r;//i=j;//}//return r;}void join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){parent[fx]=fy;//把fx的根节点赋予fy，即把城市fx的龙珠给城市移到fy num[fx]=1;//头结点移动一次 rank[fy]+=rank[fx];//根节点代表城市 }}int main(){int t;int n,m,a,b;int i;char ch;int cnt=0;scanf("%d",&t);while(t–){scanf("%d%d",&n,&m);getchar();for(i=1;i<=n;++i){parent[i]=i;num[i]=0;rank[i]=1; }//int flag=1;printf("Case %d:\n",++cnt);while(m–){scanf("%c",&ch);if(ch=='T'){scanf("%d%d",&a,&b);getchar();join(a,b);//把a所在的城市的龙珠调到城市b}else if(ch=='Q'){scanf("%d",&a);getchar();int temp=find(a); //找龙珠a在哪个城市,在找的时候不停的会加上它的父节点 //if(flag) {//printf("Case %d:",++cnt);//flag=0;//}printf("%d %d %d\n",temp,rank[temp],num[a]);} }}return 0;}
#include<stdio.h>int parent[101010];int num[101001];//表示移动次数 int rank[100010];//表示该地有多少颗龙珠 int find(int x){if(x==parent[x]) return x;int t=parent[x];parent[x]=find(parent[x]);//压缩路径 ，都指向根节点num[x]+=num[t];//每个球移动的次数等于本身移动的个数加上父节点移动的次数return parent[x]; //int r=x;//while(r!=parent[r])//{//num[r]+=num[parent[r]]; //num需要通过加上父节点来更新 //r=parent[r];//}//int i=x,j;//路径压缩 //while(i!=r)//{//j=parent[i];//parent[i]=r;//i=j;//}//return r;}void join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){parent[fx]=fy;//把fx的根节点赋予fy，即把城市fx的龙珠给城市移到fy num[fx]=1;//头结点移动一次 rank[fy]+=rank[fx];//根节点代表城市 }}int main(){int t;int n,m,a,b;int i;char ch;int cnt=0;scanf("%d",&t);while(t–){scanf("%d%d",&n,&m);getchar();for(i=1;i<=n;++i){parent[i]=i;num[i]=0;rank[i]=1; }int flag=1;//printf("Case %d:\n",++cnt);while(m–){scanf("%c",&ch);if(ch=='T'){scanf("%d%d",&a,&b);getchar();join(a,b);//把a所在的城市的龙珠调到城市b}else if(ch=='Q'){scanf("%d",&a);getchar();int temp=find(a); //找龙珠a在哪个城市,在找的时候不停的会加上它的父节点if(flag) {printf("Case %d:\n",++cnt);flag=0;}printf("%d %d %d\n",temp,rank[temp],num[a]);} }}return 0;}
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