题目描述
You are given a tree of $ n $ vertices. The vertices are numbered from $ 1 $ to $ n $ .
You will need to assign a weight to each edge. Let the weight of the $ i $ -th edge be $ a_i $ ( $ 1 \leq i \leq n-1 $ ). The weight of each edge should be an integer between $ 0 $ and $ 2^{30}-1 $ , inclusive.
You are given $ q $ conditions. Each condition consists of three integers $ u $ , $ v $ , and $ x $ . This means that the bitwise XOR of all edges on the shortest path from $ u $ to $ v $ should be $ x $ .
Find out if there exist $ a_1, a_2, \ldots, a_{n-1} $ that satisfy the given conditions. If yes, print a solution such that $ a_1 \oplus a_2 \oplus \ldots \oplus a_{n-1} $ is the smallest. Here, $ \oplus $ denotes the bitwise XOR operation.
If there are multiple solutions such that $ a_1 \oplus a_2 \oplus \ldots \oplus a_{n-1} $ is the smallest, print any.
输入格式
The first line contains two integers $ n $ and $ q $ ( $ 2 \le n \le 2.5 \cdot 10^5 $ , $ 0 \le q \le 2.5 \cdot 10^5 $ ).
The $ i $ -th of the following $ n-1 $ lines contains two integers $ x_i $ and $ y_i $ ( $ 1 \le x_i, y_i \le n $ , $ x_i \neq y_i $ ), meaning that the $ i $ -th edge connects vertices $ x_i $ and $ y_i $ in the tree.
It is guaranteed that the given edges form a tree.
The following $ q $ lines contain information about conditions.
Each line contains three integers $ u $ , $ v $ , $ x $ ( $ 1 \le u, v \le n $ , $ u \neq v $ , $ 0 \le x \le 2^{30}-1 $ ), meaning that the bitwise XOR of all edges on the shortest path from $ u $ to $ v $ should be $ x $ .
输出格式
If there do not exist $ a_1 $ , $ a_2 $ , ..., $ a_{n-1} $ that satisfy the given conditions, print "No".
Otherwise, print "Yes" in the first line.
Then print $ n-1 $ integers on the next line, where the $ i $ -th integer is the weight of the $ i $ -th edge. If there are multiple solutions that satisfy the given conditions, print a solution such that $ a_1 \oplus a_2 \oplus \ldots \oplus a_{n-1} $ is the smallest.
If there are multiple solutions such that $ a_1 \oplus a_2 \oplus \ldots \oplus a_{n-1} $ is the smallest, print any.
When printing "Yes" or "No", you can print each letter in any case (either upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.
样例 #1
样例输入 #1
4 4
1 2
2 3
3 4
1 4 3
2 4 2
1 3 1
2 3 1
样例输出 #1
No
样例 #2
样例输入 #2
6 2
1 2
2 3
3 4
2 5
5 6
1 4 2
2 6 7
样例输出 #2
Yes
4 2 4 1 6
样例 #3
样例输入 #3
6 2
1 2
2 3
3 4
2 5
5 6
1 4 3
1 6 5
样例输出 #3
Yes
6 1 4 3 0
提示
For the first example, there does not exist a set of edge weights that satisfies the given conditions.
For the second example, the two given conditions are $ a_1 \oplus a_2 \oplus a_3=2 $ and $ a_4 \oplus a_5=7 $ . There can be multiple solutions, for example, $ (a_1, a_2, a_3, a_4, a_5)=(1, 2, 1, 4, 3) $ .
For the third example, the two given conditions are $ a_1 \oplus a_2 \oplus a_3=3 $ and $ a_1 \oplus a_4 \oplus a_5=5 $ . There are multiple solutions that satisfy the given conditions.
$ (a_1, a_2, a_3, a_4, a_5)=(1, 1, 3, 4, 0) $ satisfies the given conditions, but the bitwise XOR of all edge weights is $ 7 $ , which does not have the smallest $ a_1 \oplus a_2 \oplus \ldots \oplus a_{n-1} $ , so it cannot be the answer.
在树上路径的异或值有一个很巧妙的结论：设一个根，如果根节点到点 \(x\) 的异或和为 \(v_x\)，那么路径 \((x,y)\) 的异或和为 \(v_x\oplus v_y\)
那么此时一个限制的意思就是 \(v_x\oplus v_y=a\)，转化一下，\(v_x\oplus a=v_y\)，那么此时可以通过构图的方式判定是否有解。
至于使所有点的权值异或和最小。那么考虑一个 \(v_x\) 什么时候会被算到总的点权异或和里面，当且仅当 \(x\) 在树中的度数是奇数。
那么此时要让异或和最小，可以将上面判断是否有解的图的每一个连通块钦定一个点 \(a\) 为代表，那么连通块中的每一个点的 \(v\) 值可以写为 \(v_a\oplus w\)，那么如果这一个点会计算入总点权异或和，答案一定会多异或上一个 \(w\)，\(v_a\) 是否会被计入也会发生改变。
然后此时如果没有一个连通块的点权会被计入答案，那就随便定 \(v_a\)，不影响答案。否则可以记录所有的点的 \(w\) 异或之和为 \(s\)，将一个被记入点权的 \(a\) 赋值为 \(s\)，此时总异或和为0。
// LUOGU_RID: 103370522
#include<bits/stdc++.h>
using namespace std;
const int N=3e5+5;
int n,q,d[N],in[N],f[N],p[N],ans,fl,a[N],b[N];
struct graph{
int hd[N],e_num;
struct edge{
int v,nxt,w;
}e[N<<1];
void add_edge(int u,int v,int w)
{
e[++e_num]=(edge){v,hd[u],w};
hd[u]=e_num;
// printf("%d %d %d\n",u,v,w);
}
}g,h;
void dfs(int x,int w,int t)
{
if(d[x]^w&&f[x])
{
puts("No");
exit(0);
}
if(!f[x])
{
d[x]=w,f[x]=t;
// printf("%d %d\n",x,f[x]);
for(int i=h.hd[x];i;i=h.e[i].nxt)
{
// printf("qzmyyds:%d\n",e[i].v);
dfs(h.e[i].v,w^h.e[i].w,t);
}
}
}
void sou(int x,int y)
{
for(int i=g.hd[x];i;i=g.e[i].nxt)
{
if(g.e[i].v^y)
sou(g.e[i].v,x);
else
b[g.e[i].w]=a[x]^a[y];
}
}
int main()
{
// memset(d,-1,sizeof(d));
scanf("%d%d",&n,&q);
for(int i=1;i<n;++i)
{
int u,v;
scanf("%d%d",&u,&v);
g.add_edge(u,v,i);
g.add_edge(v,u,i);
++in[u],++in[v];
}
while(q--)
{
int u,v,x;
scanf("%d%d%d",&u,&v,&x);
// printf("%d\n",x);
h.add_edge(u,v,x);
h.add_edge(v,u,x);
}
for(int i=1;i<=n;i++)
if(!f[i])
dfs(i,0,i);
// for(int i=1;i<=n;i++)
// printf("%d %d\n",f[i],d[i]);
puts("Yes");
for(int i=1;i<=n;i++)
if(in[i]&1)
p[f[i]]^=1,ans^=d[i];
// for(int i=1;i<=n;i++)
// printf("%d ",p[i]);
// puts("") ;
p[1]=0;
for(int i=2;i<=n;i++)
{
if(f[i]==i&&p[i])
p[i]=ans,ans=0;
// printf("hjhyyds:%d %d\n",d[i],p[f[i]]);
a[i]=d[i]^p[f[i]];
}
// for(int i=1;i<=n;i++)
// print
sou(1,0);
for(int i=1;i<n;i++)
printf("%d ",b[i]);
return 0;
}