hdoj--5606--tree(并查集)
tree
n
points and n−1
edges),the points are labeled from 1 to n
edge has a weight from 0 to 1,for every point i∈[1,n]
should find the number of the points which are closest to it,the clostest points can contain
i
itself.
for each test case,the first line is a nubmer n
the number of the points,next n-1 lines,each line contains three numbers
u,v,w
shows an edge and its weight.
T≤50,n≤10
in consideration of the large output,imagine ans
is the answer to point i
only need to output,ans
1
3
1 2 0
2 3 1
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.
#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\n",ans);
}
return 0;
}
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