tree

 

 Accepts: 156

 

 Submissions: 807



 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)




Problem Description



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



Input



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

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

T\le 50,n\le 10^5,u,v\in[1,n],w\in[0,1]T≤50,n≤10​5​​,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 ans_ians​i​​ is the answer to point ii,you only need to output,ans_1~xor~ans_2~xor~ans_3..~ans_nans​1​​ xor ans​2​​ xor ans​3​​.. ans​n​​.



Sample Input



1
3
1 2 0
2 3 1



Sample Output



1

in the sample.

ans_1=2ans1=2

ans_2=2ans2=2

ans_3=1ans3=1

2~xor~2~xor~1=12 xor 2 xor 1=1,so you need to output 1.
题解：找每个点距离自己最近的点的个数的异或值，注意最近点包括自己；
思路：并差集，将权值为0的点组成一颗树，这棵树代表的就是距离自己最近的点的个数；
代码：



#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define  PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const int MAXN=100010;
int a[MAXN];
int n;
struct Node{
  int u,v,w;
  Node init(){SI(u);SI(v);SI(w);}
}dt[MAXN];
int pre[MAXN],num[MAXN];
int find(int x){
  return pre[x]=x==pre[x]?x:find(pre[x]);
}
void merge(Node a){
  int f1=find(a.u),f2=find(a.v);
  if(pre[f1]!=f2)pre[f2]=f1,num[f1]+=num[f2];
}
void initial(){
  for(int i=1;i<=n;i++)pre[i]=i,num[i]=1;
}
int main(){
  int T;
  SI(T);
  while(T--){
    mem(a,0);
    SI(n);
    initial();
    int u,v,w;
    for(int i=0;i<n-1;i++){
      dt[i].init();
      w=dt[i].w;
      if(!w)merge(dt[i]);
    }int ans=0;
    for(int i=1;i<=n;i++)ans^=num[pre[i]];
    printf("%d\n",ans);
  }
  return 0;
}