More is better
http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1003&cid=12467&hide=0
Time Limit : 5000/1000ms (Java/Other) Memory Limit : 327680/102400K (Java/Other)
Total Submission(s) : 17 Accepted Submission(s) : 4
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Problem Description
Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are
still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.
Input
≠ B, 1 ≤ A, B ≤ 10000000)
Output
Sample Input
4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8
Sample Output
4 2
Hint
then A and C are also friends(indirect).
In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.
Author
Source
#include<iostream>
#include<algorithm>
using namespace std;
#define N 100005
int parent[N+5];
int height[N+5];
void init(){
for(int i=1;i<=N;i++){
parent[i]=i;
height[i]=1;
}
}
int find(int x){
while(x!=parent[x])
x=parent[x];
return x;
}
void merge(int x,int y,int *max){
x=find(x);
y=find(y);
if(x==y)
return ;
else if(height[x]==height[y]){
parent[y]=x;
height[x]+=height[y];
}
else if(height[x]<height[y]){
parent[x]=y;
height[y]+=height[x];
}
else{
parent[y]=x;
height[x]+=height[y];
}
if(*max<(height[x]>height[y]?height[x]:height[y]))
*max=(height[x]>height[y]?height[x]:height[y]);
}
int main(){
int max=1;
int n,a,b;
while( scanf("%d",&n)!=EOF){
init();
max=1;
while(n--){
scanf("%d %d",&a,&b);
merge(a,b,&max);
}
printf("%d\n",max);
}
return 0;
}