tree(并查集)

tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 986    Accepted Submission(s): 452

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5606

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

1
3
1 2 0
2 3 1

 

 

题解： 把每条边权是1的边断开，发现每个离他最近的点个数就是他所在的联通快的大小，开一个并查集，每次读到编圈是0的边就合并，最后结果就是每个点的父亲的孩子个数加1

太久没有打并查集了水一水

 #include<cstdio>
 #include<cstring>
 #include<string>
 using namespace std;
 #define N 100006
 int n;
 int fa[N];
 int num[N];
 void init(){
   for(int i = ; i < N; i++) fa[i] = i;
 }
 int find(int x){
   return fa[x]==x?x:fa[x]=find(fa[x]);
 }
 void merge(int x, int y){
   int root1 = find(x);
   int root2 = find(y);
   if(root1==root2) return;
   fa[root1] = root2;
 }

 int main()
 {
   int t;
   scanf("%d",&t);
   while(t--)
   {
     init();
     scanf("%d",&n);
     for(int i = ; i < n-; i++)
     {
       int a, b, c;
       scanf("%d%d%d",&a,&b,&c);
       if(c==){
         merge(a,b);
       }
     }
     memset(num,,sizeof(num));
     for(int i = ; i <= n; i++)
     {
       int r = find(i);
       num[r]++;
     }
     int ans = ;
     for(int i = ; i <= n; i++)
     {
       ans = (ans^num[find(i)]);
     }
     printf("%d\n",ans);
   }
   return ;
 }