Description
Farmer John has installed a new system of N?1 pipes to transport milk between the N stalls in his ba
rn (2≤N≤50,000), conveniently numbered 1…N. Each pipe connects a pair of stalls, and all stalls a
re connected to each-other via paths of pipes.FJ is pumping milk between KK pairs of stalls (1≤K≤1
00,000). For the iith such pair, you are told two stalls sisi and titi, endpoints of a path along wh
ich milk is being pumped at a unit rate. FJ is concerned that some stalls might end up overwhelmed w
ith 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 bein
g pumped through any stall. If milk is being pumped along a path from sisi to titi, then it counts a
s being pumped through the endpoint stalls sisi and titi, as well as through every stall along the p
ath between them.
给定一棵有N个点的树，所有节点的权值都为0。
有K次操作，每次指定两个点s,t，将s到t路径上所有点的权值都加一。
请输出K次操作完毕后权值最大的那个点的权值。
Input
The first line of the input contains NN and KK.
The next N-1 lines each contain two integers x and y (x≠y，x≠y) describing a pipe between stalls x and y.
The next K lines each contain two integers ss and t describing the endpoint stalls of a path through which milk is being pumped.
Output
An integer specifying the maximum amount of milk pumped through any stall in the barn.
Sample Input
5 10 3 4 1 5 4 2 5 4 5 4 5 4 3 5 4 3 4 3 1 3 3 5 5 4 1 5 3 4
Sample Output
9
树上差分，这是一道模板题，思路和普通的差分是一样的，就是需要在一棵树上进行差分的操作
那么，肯定的是s和t到他们的最近公共祖先的路径上的点的边权都是要加1的，那么就可以做
sum[s]++,sum[t]++,sum[lca(s,t)]--;
但是如果lca(s,t)不是根节点那么sum[father[lca(s,t)]]--,因为我们在加的时候s和t一共加了
两遍，所以需要把sum[lca(s,t)]--,把sum[father[lca(s,t)]]--就不用我说了吧，学过差分的都知道，如果没学过就没必要来看这个了。
代码：
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 int son[100001],nex[100001],ans,pre[100001],fa[100001],f[100001][20],n,m,a,b,root,tot,sum[100001],father[100001],now[100001],deep[100001];
5 void add(int x,int y)
6 {
7 tot++;
8 son[tot]=y;
9 nex[tot]=pre[x];
10 pre[x]=tot;
11 }
12 void dfs(int now,int fa)
13 {
14 father[now]=fa;
15 for(int i=1;i<=19;i++)
16 {
17 if(deep[now]<(1<<i))break;
18 f[now][i]=f[f[now][i-1]][i-1];
19 }
20 for(int i=pre[now];i;i=nex[i])
21 if(son[i]!=fa)
22 {
23 int k=son[i];
24 deep[k]=deep[now]+1;
25 f[k][0]=now;
26 dfs(k,now);
27 }
28 }
29 int lca(int x,int y)
30 {
31 if(deep[x]>deep[y])swap(x,y);
32 int poor=deep[y]-deep[x];
33 for(int i=0;i<=19;i++)
34 if((1<<i)&poor)
35 y=f[y][i];
36 for(int i=19;i>=0;i--)
37 if(f[x][i]!=f[y][i])
38 x=f[x][i],y=f[y][i];
39 if(x==y)return x;
40 return f[x][0];
41 }
42 void get(int x,int fa)
43 {
44 now[x]=sum[x];
45 for(int i=pre[x];i;i=nex[i])
46 if(son[i]!=fa)
47 {
48 get(son[i],x);
49 now[x]+=now[son[i]];
50 }
51 ans=max(ans,now[x]);
52 }
53 int main()
54 {
55 scanf("%d%d",&n,&m);
56 for(int i=1;i<n;i++)
57 scanf("%d%d",&a,&b),add(a,b),add(b,a);
58 dfs(1,0);
59 int begin,end;
60 for(int i=1;i<=m;i++)
61 {
62 scanf("%d%d",&begin,&end);
63 int r=lca(begin,end);
64 sum[begin]++,sum[end]++,sum[r]--;
65 if(r!=1)sum[father[r]]--;
66 }
67 get(1,0);
68 printf("%d",ans);
69 }
原文地址：https://www.cnblogs.com/lcxer/p/9441773.html