Dragon Balls
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1544 Accepted Submission(s): 627
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.

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)
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
Author
possessor WC
Source
Recommend
lcy
这题在比赛的时候一团糟,主要是球和城市把我搞晕了。比赛后才慢慢的理清思路。哎
考察点:并查集
/***************************************************************
> File Name: hudD.c
> Author: SDUT_GYX
> Mail: 2272902662@qq.com
> Created Time: 2013/5/10 9:52:40
**************************************************************/
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define N 10010
int time[N],pt[N],sum[N],tag[N],res3;
int main()
{
// freopen("data1.in","r",stdin);
int root(int x);
int i,n,m,t,T,x,y,k1,k2,res1,res2;
char c;
scanf("%d",&t);
T=1;
while(t--)
{
scanf("%d %d%*c",&n,&m);
for(i=1;i<=n;i++)
{
pt[i]=i;
time[i]=0;
sum[i]=1;
tag[i]=i;
}
printf("Case %d:\n",T++);
while(m--)
{
scanf("%c",&c);
if(c=='T')
{
scanf("%d %d%*c",&x,&y);
k1=root(x);
k2=root(y);
sum[tag[k2]]+=sum[tag[k1]];
sum[tag[k1]]=0;
tag[k1]=tag[k2];
time[k1]+=1;
pt[k1]=k2;
}else if(c=='Q')
{
scanf("%d%*c",&x);
k1=root(x);
res1=tag[k1];
res2=sum[res1];
printf("%d %d %d\n",res1,res2,res3);
}
}
}
return 0;
}
int root(int x)
{
int t1,t2,s;
t1=t2=x;
res3=0;
while(t1!=pt[t1])
{
res3+=time[t1];
t1 = pt[t1];
}
s=res3;
s-=time[t1];
while(x!=pt[x])
{
s-=time[x];
time[x]+=s;
t2=pt[x];
pt[x]=t1;
x=t2;
}
return t1;
}