F - Dragon Balls
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status
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 有n个龙珠，在1到n个城市中，T a b 移动a所在城市的所有龙珠到b所在的城市，Q a 输出a所在的城市，城市的龙珠数量，a移动的次数，并查集问题，一个数组记录每个龙珠移动的次数，一个记录城市龙珠的数量，T a b 代表a移动的次数+1，并且所有指向a所在的城市的龙珠移动次数+1 ， b所在的城市龙珠数量加a所在的城市
#include <cstdio>
#include <cstring>
int p[11000] , c[11000] , num[11000] ;
/*int f(int x)
{
int r , k , l , temp = 0 ;
r = x ;
while( r != p[r] )
{
r = p[r] ;
temp++ ;
}
k = x ;
while( k != r )
{
l = p[k] ;
c[k] += ( --temp ) ;
p[k] = r ;
k = l ;
}
return r ;
}*/
int f(int x)
{
if(x==p[x])
return x;
int t=p[x];
p[x] = f(p[x]);
c[x] += c[t];
return p[x];
}
void add(int u,int v)
{
u = f(u) ;
v = f(v) ;
if(u != v)
{
p[u] = v ;
num[v] += num[u] ;
num[u] = 0 ;
c[u]++ ;
}
}
int main()
{
int t , tt , i , n , m , a , b ;
char ch ;
scanf("%d", &t);
for(tt = 1 ; tt <= t ; tt++)
{
printf("Case %d:\n", tt);
scanf("%d %d", &n, &m);
for(i = 1 ; i <= n ; i++)
{
p[i] = i ;
num[i] = 1 ;
}
memset(c,0,sizeof(c));
while(m--)
{
scanf("%*c%c", &ch);
if( ch == 'T' )
{
scanf("%d %d", &a, &b);
add(a,b);
}
else
{
scanf("%d", &a);
int k = f(a) ;
printf("%d %d %d\n", k, num[k] , c[a] );
}
}
}
return 0;
}
测试赛F - Dragon Balls(并查集)