题目连接：
http://acm.hdu.edu.cn/showproblem.php?pid=5682
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
题意
zxa有一棵含有\(n\)个节点的无根树，包含\((n-1)\)条无向边，点从\(1\)到\(n\)编号，定义每个点的度数为与这个点相连的边的数量，度数为\(1\)的节点被称作这棵树的叶子节点。
zxa想给每个节点设置它的好看度，好看度必须为正整数。他的无根树有\(m(1\leq m\leq n)\)个叶子节点，其中的\(k(1\leq k\leq m)\)个叶子节点的好看度已经确定，zxa只需要设置其他节点的好看度。
zxa很好奇，如果令每条边的难看度是这条边所连接的两个节点的好看度差值的绝对值，整棵树的难看度是所有边的难看度中的最大值，那么这棵树的难看度最小是多少，你能帮助他吗？
题解：
二分答案之后，树形dp去跑每个节点的取值范围就好了
如果范围是允许的，那就可以直接返回true
否则就返回false
挺好的一道题
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e4+6;
int w[maxn],n,k,l[maxn],r[maxn],vis[maxn];
vector<int>E[maxn];
int dfsl(int x,int d)
{
for(int i=0;i<E[x].size();i++)
{
if(!vis[E[x][i]])
{
vis[E[x][i]]=1;
l[x]=max(l[x],dfsl(E[x][i],d));
}
}
return l[x]-d;
}
int dfsr(int x,int d)
{
for(int i=0;i<E[x].size();i++)
{
if(!vis[E[x][i]])
{
vis[E[x][i]]=1;
r[x]=min(r[x],dfsr(E[x][i],d));
}
}
return r[x]+d;
}
bool check(int mid)
{
for(int i=1;i<=n;i++)
if(w[i])l[i]=r[i]=w[i];
else l[i]=0,r[i]=1e9;
memset(vis,0,sizeof(vis));
dfsl(1,mid);
memset(vis,0,sizeof(vis));
dfsr(1,mid);
for(int i=1;i<=n;i++)
if(r[i]<l[i])return false;
return true;
}
void solve()
{
for(int i=0;i<maxn;i++)E[i].clear();
for(int i=0;i<maxn;i++)w[i]=0;
scanf("%d%d",&n,&k);
for(int i=1;i<n;i++)
{
int x,y;scanf("%d%d",&x,&y);
E[x].push_back(y);
E[y].push_back(x);
}
for(int i=1;i<=k;i++)
{
int x,y;
scanf("%d%d",&x,&y);
w[x]=y;
}
int L=0,R=1e9,ans=1e9;
while(L<=R)
{
int mid=(L+R)/2;
if(check(mid))R=mid-1,ans=mid;
else L=mid+1;
}
cout<<ans<<endl;
}
int main()
{
int t;scanf("%d",&t);
while(t--)solve();
return 0;
}
