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最新推荐文章于 2017-07-26 15:15:41 发布

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Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 8 Accepted Submission(s) : 6
Problem Description
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 $i\in[1,n]$ ,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. $T\le 50,n\le 10^5,u,v\in[1,n],w\in[0,1]$

Output
for each test case,you need to print the answer to each point. in consideration of the large output,imagine $ans_i$ is the answer to point $i$ ,you only need to output, $ans_1~xor~ans_2~xor~ans_3..~ans_n$ .

Sample Input
1
3
1 2 0
2 3 1

Sample Output
1

in the sample.

$ans_1=2$

$ans_2=2$

$ans_3=1$

$2~xor~2~xor~1=1$ ,so you need to output 1.

给你n个点的联通，让你找出与每个点的最短连接点，连接点包括它自己，然后将每个点的最短连接点个数进行或非运算
注意如果有多个点之间w都为0，我们均要将其作为一个集合，因此
直接对每个点进行处理，不过我们在rank值时变为1，然后只要当w=0时就进行并查，最后统计所有集合的点然后或非操作

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#define N 100010
using namespace std;
int map[N];
int rank_[N];
int n;
void init(int n)
{
    int i;
    for(i=1; i<=n; i++)
        map[i]=i,rank_[i]=1;
}
int find(int x)
{
    if(x!=map[x])
        map[x]=find(map[x]);
    return map[x];
}
void Union(int x,int y)
{
    x=find(x),y=find(y);
    if(x!=y)
    {
        map[x]=y;
        rank_[y]+=rank_[x];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init(n);
        int u,v,w;
        int sum=0;
        for(int i=1; i<=n-1; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            if(w==0)
                Union(u,v);
        }
        for(int i=1; i<=n; i++)
        {
            if(map[i]==i&&rank_[i]%2==1)
                sum=rank_[i]^sum;
        }
        cout<<sum<<endl;
    }
    return 0;
}