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;
}