hdu--5606（简单并查集）

tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 586    Accepted Submission(s): 286

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
   
   
   
   

 

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.
   
   
   
   

 

解题思路:这道题可谓是WA啊,WA啊.第一次写,当输入时边的权值为0，则该边两个端点记录num[ ]自增一次（记录与之距离为0的）

的点的总数），但忽略了传递性，例如a--b=0；b--c=0；那么a--b=0;然而我并没有合并。第二次写，用的是并查集，把距离为0

的点绑在一棵树上，该树上的结点数就等于该树上距每个点距离最近的点的数目，然而忘了可能最后会构成多种情况，例如，四个点，边2--4=0，1--2=1，1--3=0，这里2，4在一棵树上，1--3在一棵树上，而不是按我写的2，4，1，3在同一棵树上。再改了改，终于AC了。

代码如下：

#include<stdio.h>
int per[100000+10];
int power[100000+10];
void inin(int n){
	for(int i=1;i<=n;i++){
		per[i]=i;
		power[i]=1;
	}
}
int find(int x){
	return per[x]=x==per[x]?x:find(per[x]);
}
void judge(int x,int y){
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy){
		per[fx]=fy;
		power[fy]+=power[fx];
	}
}
int main(){
	int t,n,a,b,c;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		inin(n);
		for(int i=1;i<n;i++){
			scanf("%d%d%d",&a,&b,&c);
			if(c==0){
				judge(a,b);
			}
		}
		int ans=power[find(1)];
		for(int i=2;i<=n;i++){
			ans^=power[find(i)];
		}
		printf("%d\n",ans);
	}
	return 0;
}