HDOJ 5606-tree【并查集】

tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 894    Accepted Submission(s): 420

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)

解题思路：

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)。

由题意可得，当权值为1时这两边是不会联通的，因为这一点到自己的距离始终唯0，所以这一点只有到距离为0常能联通。

#include<stdio.h>#include<string.h>#include<algorithm>#define M 100002using namespace std;int n;int fa[M],arr[M];void init(){int i,j;for(i=1;i<=n;i++){fa[i]=i;arr[i]=1;}}int find(int x){int r=x;while(r!=fa[r]){r=fa[r];}return r;}void join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){fa[fx]=fy;arr[fy]+=arr[fx];}}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);init();int i,j;for(i=0;i<n-1;i++){int x,y,w;scanf("%d%d%d",&x,&y,&w);if(w==1){continue;}else{join(x,y);}}int ans=arr[find(1)];for(i=2;i<=n;i++){ans^=arr[find(i)];//异或运算 }printf("%d\n",ans);}return 0;}