hdoj--5606--tree(并查集)

tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 557    Accepted Submission(s): 271

Problem Description

There is a tree(the tree is a connected graph which contains n points and n−1 edges),the points are labeled from 1 to n,which edge has a weight from 0 to 1,for every point i∈[1,n],you should find the number of the points which are closest to it,the clostest points can contain i itself.

 

Input

the first line contains a number T,means T test cases.

for each test case,the first line is a nubmer n,means the number of the points,next n-1 lines,each line contains three numbers u,v,w,which shows an edge and its weight.

T≤50,n≤105,u,v∈[1,n],w∈[0,1]

 

Output

for each test case,you need to print the answer to each point.

in consideration of the large output,imagine ansi is the answer to point i,you only need to output,ans1 xor ans2 xor ans3.. ansn.

 

Sample Input

1
3
1 2 0
2 3 1

 

Sample Output

1

in the sample.

$ans_1=2$

$ans_2=2$

$ans_3=1$

$2~xor~2~xor~1=1$,so you need to output 1.

 

Source

BestCoder Round #68 (div.2)

 

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借鉴大神的思路，太牛！

递归一直出错，最后乖乖用了循环，是不是压缩路径的方式不对啊，

#include<stdio.h>
#include<string.h>
int rank[100100];
int pre[100100];
void init()
{
	for(int i=0;i<100100;i++)
	pre[i]=i;
}
//int find(int x)
//{
//	return pre[x]==x?x:pre[x]=find(pre[x]);
//}
int find(int p)
{
	int child=p;
	while(p!=pre[p])
	p=pre[p];
	while(child!=p)
	{
		int t=pre[child];
		pre[child]=p;
		child=t;
	}
	return p;
}
void join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fy!=fx)
	pre[fx]=fy;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		memset(rank,0,sizeof(rank));
		init();
		int a,b,c;
		scanf("%d",&n);
		for(int i=1;i<n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(c==0)
			join(a,b);
		}
		for(int i=1;i<=n;i++)
		{
			if(pre[i]==i)
				rank[i]++;
			else
				rank[find(i)]++;
		}
		int ans=0;
		for(int i=1;i<=n;i++)
		ans^=(rank[find(i)]);
		printf("%d
",ans);
	}
	return 0;
}