HDU 5606 Tree【并查集】

最新推荐文章于 2021-10-25 19:11:05 发布

原创最新推荐文章于 2021-10-25 19:11:05 发布 · 200 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#并查集

数据结构--并查集专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种利用并查集解决树形结构中点间距离问题的方法，通过合并权重为0的边连接的节点集合，快速计算每个节点与其相同距离节点的数量。并提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
}