A Possible Tree（ICPC2017 Urumqi）

A Possible Tree

题目描述

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 W conclusions 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.

输入

The input has several test cases and the first line contains an integer t (1 ≤ t ≤ 30) which is the number of test cases.
For each case, the first line contains two integers n (1 ≤ n ≤ 100000) and c (1 ≤ c ≤ 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 v (1 ≤ u, v ≤ 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 1 ≤ u, v ≤ n and val ∈ [0, 260 − 1].

输出

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

题意：问从前面开始不冲突的个数
　　Get 新知识点：带权并查集
　　这道题就是带权并查集的模本题啊
　　注意 ^ 的优先级 小于 != 
AC code:

#include <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const int N =  1e5 + 10;
int pre[N];
ll r[N];
int n;
void init()
{
    for(int i = 1;i <= n;i++)   pre[i] = i,r[i] = 0;
}
 
int find(int x)
{
    if(x == pre[x]) return x;
    int prex = pre[x];
    pre[x] = find(prex);
    r[x] = r[x]^r[prex];
    return pre[x];
}
 
int add(int x,int y,ll val)
{
    int prex = find(x),prey = find(y);
    if(prex == prey)
    {
        if((r[x]^r[y]) != val)
            return 0;
    }
    else
    {
        pre[prex] = prey;
        r[prex] = r[x]^r[y]^val;
    }
    return 1;
}
int main()
{
 
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int m,u,v;
        ll val;
        scanf("%d%d",&n,&m);
        init();
        for(int i = 1;i < n;i++)    scanf("%d%d",&u,&v);
        int ans = m,flag = 0;
        for(int i = 0;i < m;i++)
        {
            scanf("%d%d%lld",&u,&v,&val);
            if(!flag&&!add(u,v,val))
                ans = i,flag = 1;
        }
        printf("%d
",ans);
    }
    return 0;
}