zxa and leaf

Posted 2021-02-13 czy-power

zxa and leaf

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 502    Accepted Submission(s): 198

Problem Description

zxa have an unrooted tree with n nodes, including (n−1) undirected edges, whose nodes are numbered from 1 to n . The degree of each node is defined as the number of the edges connected to it, and each node whose degree is 1 is defined as the leaf node of the tree.

zxa wanna set each node‘s beautiful level, which must be a positive integer. His unrooted tree has m(1≤m≤n) leaf nodes, k(1≤k≤m) leaf nodes of which have already been setted their beautiful levels, so that zxa only needs to set the other nodes‘ beautiful levels.

zxa is interested to know, assuming that the ugly level of each edge is defined as the absolute difference of the beautiful levels between two nodes connected by this edge, and the ugly level of the tree is the maximum of the ugly levels of **all the edges on this tree**, then what is the minimum possible ugly level of the tree, can you help him?

 

 

Input

The first line contains an positive integer T , represents there are T test cases.

For each test case:

The first line contains two positive integers n and k , represent the tree has n nodes, k leaf nodes of which have already been setted their beautiful levels.

The next (n−1) lines, each line contains two distinct positive integers u and v , repersent there is an undirected edge between node u and node v .

The next k lines, each lines contains two positive integers u and w , repersent node u is a leaf node, whose beautiful level is w .

There is a blank between each integer with no other extra space in one line.

It‘s guaranteed that the input edges constitute a tree.

1≤T≤10,2≤n≤5⋅104,1≤k≤n,1≤u,v≤n,1≤w≤109

 

 

Output

For each test case, output in one line a non-negative integer, repersents the minimum possible ugly level of the tree.

 

 

Sample Input

2 3 2 1 2 1 3 2 4 3 9 6 2 1 2 1 3 1 4 2 5 2 6 3 6 5 9

 

 

Sample Output

3 1

Hint

If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.

 

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn=1e5+10;
struct node{
     int to,nx;
}p[maxn];
int T,n,k,head[maxn],vis[maxn],tot,d[maxn];
ll small[maxn],big[maxn],val[maxn];
void addedge(int u,int v){
       p[++tot].to=v,p[tot].nx=head[u],head[u]=tot;
}
bool dfs(int now,int fa,int w){
      for(int i=head[now];i;i=p[i].nx){
            int to=p[i].to;
            if(to==fa)continue;
            if(!dfs(to,now,w))return false;
            ll tempsmall=small[to]-w;
            ll tempbig=big[to]+w;
            if(tempsmall>big[now]||tempbig<small[now])return false;
            small[now]=max(small[now],tempsmall);
            big[now]=min(big[now],tempbig);
      }
      return true;
}
bool ok(int cur,int root){
      for(int i=1;i<=n;i++){
           if(vis[i])continue;
           small[i]=-0x3f3f3f3f;
           big[i]=0x3f3f3f3f;
      }
      if(dfs(root,-1,cur))return true;
      else return false;
}
int main()
{
    scanf("%d",&T);
    while(T--){
        tot=0;
        scanf("%d%d",&n,&k);
        memset(d,0,sizeof(d));
        memset(head,0,sizeof(head));
        memset(vis,0,sizeof(vis));
        memset(small,0,sizeof(small));
        memset(big,0,sizeof(big));
        for(int i=1;i<=n-1;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            addedge(a,b);
            addedge(b,a);
            d[a]++,d[b]++;
        }
        while(k--){
            int a,b;
            scanf("%d%d",&a,&b);
            small[a]=big[a]=b;
            vis[a]=1;
        }
        int root=1;
        for(int i=1;i<=n;i++){
            if(!d[i]){
                root=i;
                break;
            }
        }
        int l=0,r=0x3f3f3f3f;
        while(l<=r){
            int mid=l+r>>1;
            if(ok(mid,root))r=mid-1;
            else l=mid+1;
        }
        cout<<l<<endl;
    }
    return 0;
}