【并查集】HDOJ tree 5606

tree

                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                             Total Submission(s): 448    Accepted Submission(s): 222

Problem Description

There is a tree(the tree is a connected graph which containsn 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 containi 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

131 2 02 3 1

 

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.

 

Source

BestCoder Round #68 (div.2)

 

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题意：

有n个点，给n-1个边和权值w，找出i点所有距离为0的点的数量ansi，XOR所有点的结果ansi。

解题思路：

并查集~把所有权值为0点的并到一个连通图上，然后遍历所有的点的根节点的ans值进行操作。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 100000+10;int ans[MAXN];int p[MAXN];void init(int n){    for(int i=1;i<=n;i++){        p[i]=i;ans[i]=0;    }}int find(int x){    int son,temp;    son=x;    while(x!=p[x]){        x=p[x];    }    while(son!=x){        temp=p[son];        p[son]=x;        son=temp;    }    return x;}void merge(int x,int y){    x=find(x);y=find(y);    if(x!=y)p[x]=y;}int main(){    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        init(n);        int u,v;        int w;        for(int i=0;i<n-1;i++){            scanf("%d%d%d",&u,&v,&w);            if(!w)merge(u,v);        }        for(int i=1;i<=n;i++){ //一定要进行一次压缩            find(i);        }        for(int i=1;i<=n;i++){            ans[p[i]]++;        }        int res=0;        for(int i=1;i<=n;i++){            res^=ans[p[i]];        }        printf("%d\n",res);    }    return 0;}