HDU 5606 Tree【并查集】

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题目：

There is a tree(the tree is a connected graph which contains nnpoints 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∈[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≤50,n≤105,u,v∈[1,n],w∈[0,1]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 ansiansi is the answer to point ii,you only need to output,ans1 xor ans2 xor ans3.. ansnans1 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.

 题目大意：

        给定n个点，n-1条带权边，边的权值为0或1，求每个点有多少与它距离为0的点，视点到自身的距离为0；

解题思路：

        并查集，当两点之间的权值为0时，合并两集合，用标记节点记录 标记节点 存在 的 集合中的元素数目。

实现代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100000+5;

//并查集
int fa[maxn],num[maxn];
int findset(int x){
    return fa[x]==x? x: fa[x]=findset(fa[x]);
}
void bin(int u,int v){
    int fu=findset(u);
    int fv=findset(v);
    if(fu!=fv){
        fa[fv]=fu;  //改变父节点为fu
        num[fu]+=num[fv];
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        for(int i=1;i<100003;i++){
            fa[i]=i;
            num[i]=1;
        }

        int n;
        scanf("%d",&n);

        for(int i=1;i<=n-1;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(w==0){
                bin(u,v);
            }
        }

        int ans=0;
        for(int i=1;i<=n;i++){  //找到每一个节点的父节点的集合
            ans^=num[findset(i)];
        }
        printf("%d\n",ans);
    }
    return 0;
}