因为徐州现场赛的G是树上差分+组合数学，但是比赛的时候没有写出来(自闭)，背锅。
会差分数组但是不会树上差分，然后就学了一下。
看了一些东西之后，对树上差分写一点个人的理解：
首先要知道在树上，两点之间只有一条路径。树上差分就是在树上用差分数组，因为是在树上的操作，所以要用到lca，因为对于两点a，b，从a到b这一条链就是a-->lca(a,b)-->b，这是一条链。
其次，树上差分的两种操作：一种是对点权的，另一种是对边权的。
对于点权：
在树上将路径的起点a+1和终点b+1，lca(a,b)-1，lca(a,b)的爸爸-1，因为对于点权，从a到b包括lca(a,b)，因为lca其实是+2，lca(a,b)多算了一次，所以lca(a,b)-1，但是lca(a,b)往上的爸爸们都没有计算到，所以lca的爸爸-1就可以。总的代码就是
sum[a]++;sum[b]++;sum[lca]--;sum[fa[lca][0]]--;
对于边权：
在树上，对于a和b，用深度更深的那个数表示边的编号，对于a和b连接的边，如果a是b的爸爸，那么这条边的编号就是b。
在树上将起点边a+1，终点边b+1，lca(a,b)-2，因为对于边权，从边a到边b，不包括lca(a,b)，因为lca(a,b)这一条边并不在从a到b的路径上，但是lca(a,b)其实多算了2次，所以要减去，就是lca(a,b)-2。
代码就是
sum[a]++;sum[b]++;sum[lca]-=2;
其他的好像也没什么了，就是dfs的时候别写捞了就可以。
先贴一个关于点权的树上差分。
P3128 [USACO15DEC]最大流Max Flow
题目描述
Farmer John has installed a new system of N-1N−1 pipes to transport milk between the NN stalls in his barn (2 leq N leq 50,0002≤N≤50,000), conveniently numbered 1 ldots N1…N. Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes.
FJ is pumping milk between KK pairs of stalls (1 leq K leq 100,0001≤K≤100,000). For the iith such pair, you are told two stalls s_isi and t_iti, endpoints of a path along which milk is being pumped at a unit rate. FJ is concerned that some stalls might end up overwhelmed with all the milk being pumped through them, since a stall can serve as a waypoint along many of the KK paths along which milk is being pumped. Please help him determine the maximum amount of milk being pumped through any stall. If milk is being pumped along a path from s_isi to t_iti, then it counts as being pumped through the endpoint stalls s_isi and
t_iti, as well as through every stall along the path between them.
FJ给他的牛棚的N(2≤N≤50,000)个隔间之间安装了N-1根管道，隔间编号从1到N。所有隔间都被管道连通了。
FJ有K(1≤K≤100,000)条运输牛奶的路线，第i条路线从隔间si运输到隔间ti。一条运输路线会给它的两个端点处的隔间以及中间途径的所有隔间带来一个单位的运输压力，你需要计算压力最大的隔间的压力是多少。
输入输出格式
输入格式：
The first line of the input contains NN and KK.
The next N-1N−1 lines each contain two integers xx and yy (x e yx≠y) describing a pipe
between stalls xx and yy.
The next KK lines each contain two integers ss and tt describing the endpoint
stalls of a path through which milk is being pumped.
输出格式：
An integer specifying the maximum amount of milk pumped through any stall in the
barn.
输入输出样例
题目就是关于点权的，树上差分的模板题。直接代码。
代码:
1 //洛谷 P3128 [USACO15DEC]最大流Max Flow -树上差分(点的树上差分) 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 #include<bitset> 7 #include<cassert> 8 #include<cctype> 9 #include<cmath> 10 #include<cstdlib> 11 #include<ctime> 12 #include<deque> 13 #include<iomanip> 14 #include<list> 15 #include<map> 16 #include<queue> 17 #include<set> 18 #include<stack> 19 #include<vector> 20 using namespace std; 21 typedef long long ll; 22 typedef pair<int,int> pii; 23 24 const double PI=acos(-1.0); 25 const double eps=1e-6; 26 const ll mod=1e9+7; 27 const int inf=0x3f3f3f3f; 28 const int maxn=1e5+10; 29 const int maxm=100+10; 30 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 31 32 struct node{ 33 int to,next; 34 }edge[maxn<<2]; 35 36 int head[maxn<<2],sum[maxn],dep[maxn],fa[maxn][30],n,m,cnt,ans; 37 38 int read()//加速挂 39 { 40 int res=0; 41 char c=getchar(); 42 while(c<'0'||c>'9') c=getchar(); 43 while(c>='0'&&c<='9') res=res*10+c-'0',c=getchar(); 44 return res; 45 } 46 47 void add(int x,int y){edge[++cnt].to=y,edge[cnt].next=head[x],head[x]=cnt;}//链式前向星存图 48 49 void dfs(int u,int fath) 50 { 51 dep[u]=dep[fath]+1,fa[u][0]=fath; 52 for (int i=0;fa[u][i];++i) fa[u][i+1]=fa[fa[u][i]][i]; 53 for (int i=head[u];i;i=edge[i].next){ 54 int v=edge[i].to; 55 if(v!=fath) dfs(v,u); 56 } 57 } 58 59 int LCA(int u,int v) 60 { 61 if(dep[u]>dep[v]) swap(u,v); 62 for (int i=20;i>=0;--i) if(dep[u]<=dep[v]-(1<<i)) v=fa[v][i]; 63 if(u==v) return u; 64 for (int i=20;i>=0;--i) if(fa[u][i]!=fa[v][i]) u=fa[u][i],v=fa[v][i]; 65 return fa[u][0]; 66 } 67 68 void Dfs(int u,int fath)//最后的遍历操作 69 { 70 for (int i=head[u];i;i=edge[i].next){ 71 int v=edge[i].to; 72 if(v==fath) continue; 73 Dfs(v,u); 74 sum[u]+=sum[v]; 75 } 76 ans=max(ans,sum[u]); 77 } 78 79 int main() 80 { 81 n=read(),m=read(); 82 int x,y; 83 for (int i=1;i<n;i++){ 84 x=read(),y=read(); 85 add(x,y);add(y,x); 86 } 87 dfs(1,0); 88 for (int i=1;i<=m;++i){ 89 x=read();y=read(); 90 int lca=LCA(x,y); 91 ++sum[x];++sum[y];--sum[lca];--sum[fa[lca][0]]; 92 } 93 Dfs(1,0); 94 printf("%d ",ans); 95 return 0; 96 }
溜了，本来昨天就改写的，忘了。