ACM-ICPC 2017 Asia Urumqi： I. A Possible Tree（带权并查集）

Alice knows that Bob has a secret tree (in terms of graph theory) with n nodes with n -1 weighted edges with integer values in $[0,260-1]$. She knows its structure but does not know the specific information about edge weights.

Thanks to the awakening of Bob’s conscience, Alice gets m conclusions related to his tree. Each conclusion provides three integers u,v and val saying that the exclusive OR (XOR) sum of edge weights in the unique shortest path between u and v is equal to val.

Some conclusions provided might be wrong and Alice wants to find the maximum number W such that the first W given conclusions are compatible. That is say that at least one allocation of edge weights satisfies the first Wconclusions all together but no way satisfies all the first W + 1 conclusions (or there are only W conclusions provided in total).

Help Alice find the exact value of W .

Input

The input has several test cases and the first line contains an integer $t(1\le t\le 30)$ which is the number of test cases.

For each case, the first line contains two integers $n(1\le n\le 100000)$ and $c(1\le c\le 100000)$ which are the number of nodes in the tree and the number of conclusions provided. Each of the following n-1 lines contains two integers u and $(1\le u,v\le n)$ indicating an edge in the tree between the u-th node and the v-th node. Each of the following c lines provides a conclusion with three integers u, v and val where $\le u$, $v\le n$ and $val\in [0,260-1]$.

Output

For each test case, output the integer W in a single line.

样例输入

2
7 5
1 2
2 3
3 4
4 5
5 6
6 7
1 3 1
3 5 0
5 7 1
1 7 1
2 3 2
7 5
1 2
1 3
1 4
3 5  
3 6
3 7  
2 6 6
4 7 7
6 7 3
5 4 5
2 5 6

样例输出

3 4

思路：带权并查集。r[i]表示i到p[i]路径上所有边的异或值。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e5+10;
const int MOD=1e9+7;
const double PI=acos(-1.0);
typedef long long ll;
int p[MAX],r[MAX];
int f(int x)
{
    if(p[x]==x)return x;
    int fa=p[x];
    p[x]=f(p[x]);
    r[x]^=r[fa];
    return p[x];
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,c;
        scanf("%d%d",&n,&c);
        for(int i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
        }
        for(int i=0;i<=n;i++)
        {
            p[i]=i;
            r[i]=0;
        }
        int ans=c+1;
        for(int i=1;i<=c;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            if(ans!=c+1)continue;
            int fx=f(x),fy=f(y);
            if(fx==fy&&(r[x]^r[y])!=z)ans=i;
            else
            {
                p[fx]=fy;
                r[fx]=r[x]^r[y]^z;
            }
        }
        printf("%d\n",ans-1);
    }
    return 0;
}