5220: A Possible Tree
时间限制: 2 Sec 内存限制: 128 MB
提交: 41 解决: 11
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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
【带权并查集】新鲜玩意。
并查集的基础上多一个数组，比如可以用sum[]多维护一个权值。
朴素的并查集模型，对于每个连通块有一个“队长”，那么就用sum[x]表示从x点到x的队长点之间的权值（比如本题就是维护异或和）。
在更新并查集时同时更新sum数组即可。
意义：对于仅确定区间权值和而不确定具体每个权值时，带权并查集是理想的选择。
【题意】
给出一棵树，然后又m条信息： x y val 表示x到y之间的异或和为val。m条信息可能存在冲突。
问最多前几条信息是不冲突的。
【分析】
对本题来说，给个树没什么jb用，无非就是想说每两点之间的路径是唯一的。我们并不关心树的形状。
带权并查集处理信息，并用sum数组维护每个点x到其本连通块队长之间的路径异或和。
当某一条信息中两点x与y在统一连通块上，那么就必须满足 “x->队长” ^ “y->队长” == val，否则就有了冲突。
【代码】
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=2e5+5;
int fa[MAX];
ll sum[MAX];
int father(int x)
{
if(x==fa[x])return x;
int fx=fa[x];
fa[x]=father(fa[x]);
sum[x]^=sum[fx]; //由于x更换了队长，换的路径要算上
return fa[x];
}
bool join(int x,int y,ll val)
{
int fx=father(x);
int fy=father(y);
if(fx==fy)return false;
sum[fx]^=sum[x]^sum[y]^val; //由于是异或，疑惑一下就抵消了
fa[fx]=fy;
return true;
}
int main()
{
int T,n,m,x,y;
ll val;
cin>>T;
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n-1;i++)scanf("%*d%*d");
for(int i=1;i<=n;i++)fa[i]=i,sum[i]=0;
int ans=m;
for(int i=1;i<=m;i++)
{
scanf("%d%d%lld",&x,&y,&val);
if(ans==m&&!join(x,y,val)&&(sum[x]^sum[y])!=val)
{
ans=i-1;
}
}
printf("%d\n",ans);
}
}