| A friend is like a flower, a rose to be exact, Or maybe like a brand new gate that never comes unlatched. A friend is like an owl, both beautiful and wise. Or perhaps a friend is like a ghost, whose spirit never dies. A friend is like a heart that goes strong until the end. Where would we be in this world if we didn't have a friend? - By Emma Guest |
Now you've grown up, it's time to make friends. The friends you make in university are the friends you make for life. You will be proud if you have many friends.
Input
There are multiple test cases for this problem.
Each test case starts with a line containing two integers N, M (1 <= N <= 100'000, 1 <= M <= 200'000), representing that there are totally N persons (indexed from 1 to N) and M operations, then M lines with the form "M a b" (without quotation) or "Q a" (without quotation) follow. The operation "M a b" means that person a and b make friends with each other, though they may be already friends, while "Q a" means a query operation.
Friendship is transitivity, which means if a and b, b and c are friends then a and c are also friends. In the initial, you have no friends except yourself, when you are freshman, you know nobody, right? So in such case you have only one friend.
Output
For each test case, output "Case #:" first where "#" is the number of the case which starts from 1, then for each query operation "Q a", output a single line with the number of person a's friends.
Separate two consecutive test cases with a blank line, but Do NOT output an extra blank line after the last one.
Sample Input
3 5
M 1 2
Q 1
Q 3
M 2 3
Q 2
5 10
M 3 2
Q 4
M 1 2
Q 4
M 3 2
Q 1
M 3 1
Q 5
M 4 2
Q 4
Sample Output
Case 1:
2
1
3
Case 2:
1
1
3
1
4
Notes
This problem has huge input and output data, please use 'scanf()' and 'printf()' instead of 'cin' and 'cout' to avoid time limit exceed.
//算是并查集模板题了
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#define Y 100003
using namespace std;
int f[Y],r[Y],h[Y];
int Find(int x)
{
if(x!=f[x])
return f[x]=Find(f[x]);
return x;
}
void union_set(int x,int y)
{
x=Find(x);
y=Find(y);
if(x==y) return ;
if(r[x]>r[y])
f[y]=x,h[x]+=h[y];
else if(r[x]==r[y])
r[x]++,f[y]=x,h[x]+=h[y];
else
f[x]=y,h[y]+=h[x];
}
int main()
{
int N,M,t=1;
bool b=1;
char c;
int x,y;
while(scanf("%d%d",&N,&M)!=EOF)
{
if(b) b=0;else printf("\n");
printf("Case %d:\n",t++);
for(int i=1;i<=N;i++)
f[i]=i,r[i]=h[i]=1;
while(M--)
{
getchar();
scanf("%c",&c);
if(c=='M')
{
scanf("%d%d",&x,&y);
union_set(x,y);
}
else
{
scanf("%d",&x);
x=Find(x);
printf("%d\n",h[x]);
}
}
}
return 0;
}
本文介绍了一个关于友谊关系的问题,通过编程解决如何判断人物之间的间接友谊关系,并提供了完整的代码实现。
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