hdoj 5606 tree 【并查集】

最新推荐文章于 2020-09-18 08:46:11 发布

原创最新推荐文章于 2020-09-18 08:46:11 发布 · 453 阅读

0 ·

CC 4.0 BY-SA版权

并查集专栏收录该内容

35 篇文章

订阅专栏

tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 180 Accepted Submission(s): 95

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∈[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

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.

问题描述

有一个树( $n$ 个点,  $n - 1$ 条边的联通图),点标号从 $1$ ~ $n$ ,树的边权是 $0$ 或 $1$ .求离每个点最近的点个数(包括自己).

输入描述

第一行一个数字 $T$ ,表示 $T$ 组数据.
对于每组数据,第一行是一个 $n$ ,表示点个数,接下来 $n - 1$ ,每行三个整数 $u, v, w$ ,表示一条边连接的两个点和边权.
 $\leq n \leq 100000, 1 \leq u,v \leq n,0 \leq w \leq 1$ .

输出描述

对于每组数据,输出答案.
考虑到输出规模过大,设 $ans_i$ 表示第 $i$ 个点的答案.你只需输出 $ans_1 \ xor \ ans_2 \ xor \ ans_3.. \ xor \ ans_n$ 即可.

输入样例

输出样例

Hint

 $ans_1 = 2$ 
 $ans_2 = 2$ 
 $ans_3 = 1$ 
 $\ xor \ 2 \ xor \ 1=1$ , 因此输出 $1$ .

思路：并查集把距离为0的点合并在一起，然后每次查找根节点就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (500000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
int ans[MAXN];
int pre[MAXN];
int Find(int p)
{
    int t;
    int child = p;
    while(p != pre[p])
        p = pre[p];
    while(child != p)
    {
        t = pre[child];
        pre[child] = p;
        child = t;
    }
    return p;
}
void Merge(int x, int y)
{
    int fx = Find(x);
    int fy = Find(y);
    if(fx != fy)
        pre[fx] =fy;
}
int main()
{
    int t; Ri(t);
    W(t)
    {
        int n; Ri(n);
        CLR(ans, 0);
        for(int i = 1; i <= n; i++)
            pre[i] = i;
        for(int i = 0; i < n-1; i++)
        {
            int a, b, c;
            Ri(a); Ri(b); Ri(c);
            if(c == 0)
                Merge(a, b);
        }
        for(int i = 1; i <= n; i++)
        {
            //printf("%d\n", pre[i]);
            if(pre[i] == i)
                ans[i]++;
            else
                ans[Find(i)]++;
        }
        int Ans = 0;
        for(int i = 1; i <= n; i++)
            Ans ^= (ans[Find(i)]);// Pi(ans[Find(i)]);
        Pi(Ans);
    }
    return 0;
}