hdu5606 tree 并查集

最新推荐文章于 2019-01-19 12:07:41 发布

moon548834

最新推荐文章于 2019-01-19 12:07:41 发布

阅读量248

点赞数

分类专栏：并查集文章标签： ACM

本文链接：https://blog.youkuaiyun.com/moon548834/article/details/59443053

版权

并查集专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一种基于树形结构的算法问题，探讨了如何通过遍历和连接权重为0的节点来确定每个节点所属联通块的个数，并最终计算这些个数的异或值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

itself.

Input

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

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

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 ansi is the answer to point i ,you only need to output, ans1 xor ans2 xor ans3.. ansn

Sample Input

Sample Output

     1in the sample.$ans_1=2$$ans_2=2$$ans_3=1$$2~xor~2~xor~1=1$,so you need to output 1. 
   

题面：给你一些点，求各个点联通块的个数的异或值。遇到1则断，0是联通。

比较容易，0就unit，1就不管好了,最后输出的时候整理一下fa[i]的值如果是偶数就跳过，因为偶数次异或是0，遇到奇数则操作。

比如sample中其实不需要算2^2，只需要算1就好，同理如果有2个2,3个1的话，只需要考虑1，而不需要考虑1^1^1。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <cctype>
#include <ctime>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
#define N 100050
#define eps 1e-5
#define pai acos(-1)
#define MOD 1e9+7
int fa[N],a[N];
void init()
{
    for(int i=1;i<=100005;i++)
    {
        fa[i]=i;
        a[i]=0;
    }
}
int findd(int x)
{
    if(x==fa[x]) return x;
    else return fa[x]=findd(fa[x]);
}
void unite(int x,int y)
{
    x=findd(x),y=findd(y);
    if(x==y) return ;
    else fa[x]=y;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    ll _begin=clock();
    #endif
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        init();
        scanf("%d",&m);
        int mm=m;
        while(--m)
        {
            int from,to,w;
            scanf("%d%d%d",&from,&to,&w);
            if(!w)
            {
                unite(from,to);
            }
        }
        m=mm;
        for(int i=1;i<=m;i++) fa[i]=findd(i);
        int ans=0;
        for(int i=1;i<=m;i++)
        {
            a[fa[i]]++;
        }
        for(int i=1;i<=100000;i++)
        {
            if(a[i]%2==0) continue;
            ans=ans^a[i];
        }
        printf("%d\n",ans);
    }
    #ifndef ONLINE_JUDGE
    ll _end=clock();
    printf("%lldms\n",_end-_begin);
    #endif

    return 0;
}