tree（简单并差集）

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 iin[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.

Tle 50,nle 10^5,u,vin[1,n],win[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=2ans​1​​=2

ans_2=2ans​2​​=2

ans_3=1ans​3​​=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
",ans);
	}
	return 0;
}