A Possible Tree

最新推荐文章于 2021-10-25 17:39:07 发布

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最新推荐文章于 2021-10-25 17:39:07 发布

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本文探讨了APossibleTree算法的实现，通过并查集处理树结构中的边权重问题，解决Alice寻找最大兼容结论数W的挑战。文章详细介绍了算法流程，包括初始化、查找与合并操作，以及如何通过输入输出样例理解问题。

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5220: A Possible Tree

时间限制: 2 Sec 内存限制: 128 MB
提交: 133 解决: 42
[提交] [状态] [讨论版] [命题人:admin]

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

样例输出

3
4

并查集，意外的不是很难。。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e5+5;
int f[maxn], r[maxn];
int n;
void init()
{
    for (int i = 0 ; i <= n ; i ++)
    {
        f[i] = i;
        r[i] = 0;
    }
}
int Find(int x)
{
    if (f[x] == x) return x;

    int tmp = f[x];
    f[x] = Find(f[x]);
    r[x] ^= r[tmp];
    return f[x];
}
bool Un(int u, int v, int s)
{
    int fu = Find(u);
    int fv = Find(v);
    if (fu != fv)
    {
        f[fu] = fv;
        r[fu] = r[u] ^ r[v] ^ s;
        return 1;
    }
    if ((r[u] ^ r[v]) != s) return 0;
    return 1;
}
void solve()
{
    int m;
    scanf("%d%d", &n, &m);
    init();
    for (int i = 1 ; i < n ; i ++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
    }
    int flag = 0;
    for (int i = 1 ; i <= m ; i ++)
    {
        int u, v, s;
        scanf("%d%d%d", &u, &v, &s);
        if (!Un(u, v, s) && !flag)
        {
            flag = i;
        }
    }
    printf("%d\n", flag - 1);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
        solve();
    return 0;
}