zoj-2833 friendship

A - Friendship
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
SubmitStatus

Description

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.

时间超限啊，这咋办

#include<stdio.h>
int bin[100005];
int find(int x)
{
int r;
r=x;
while(bin[r]!=r)
r=bin[r];
return r;
}
void  bing(int x,int y)
{   int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
{
bin[fx]=fy;

//num[fx]=find(fy);
}
}

int main()
{

int n,m,x,y,t,d,count,e=1;
char c;
while (scanf("%d%d",&n,&m)!=EOF)
{ getchar();
printf("Case %d:\n",e++);
for(int i=1;i<=n;i++)
{bin[i]=i;}
for(int i=0;i<m;i++)
{  count=0;
scanf("%c",&c);
if(c=='M')
{
scanf("%d%d",&x,&y);getchar();
bing(x,y);
}
else if (c=='Q')
{
scanf("%d",&t);getchar();
d=find(t);
for(int i=1;i<=n;i++)
if(find(i)==d)  count++;

printf("%d\n",count );
}
}

}return 0;

}

换思路，下面是通过的代码，就是计数，把每个根的数都统计出来160MS过

：

#include<stdio.h>
int bin[100005],num[100005];
int find(int x)
{
int r;
r=x;
while(bin[r]!=r)
r=bin[r];
return r;
}
void  bing(int x,int y)
{   int fx,fy;
fx=find(x);
fy=find(y);
if(fx==fy) return ;
else  if(fx<=fy)
{
bin[fx]=fy;

num[fy]+=num[fx];
}
else  if(fx>=fy)
{
bin[fy]=fx;

num[fx]+=num[fy];
}
}

int main()
{

int n,m,x,y,t,d,count,e=1,q=0;
char c;
while (scanf("%d%d",&n,&m)!=EOF)
{ getchar();
++q;
if(q!=1)
printf("\n");
printf("Case %d:\n",e++);
for(int i=1;i<=n;i++)
{bin[i]=i;num[i]=1;}
for(int i=0;i<m;i++)
{  count=0;
scanf("%c",&c);
if(c=='M')
{
scanf("%d%d",&x,&y);getchar();
bing(x,y);
}
else if (c=='Q')
{
scanf("%d",&t);

printf("%d\n",num[find(t)]) ;getchar();
}

}

}
return 0;

}