【题目】

传送门

题目描述：

Farmer John has installed a new system of N-1 pipes to transport milk between the N stalls in his barn (2 ≤ N ≤ 50,000), conveniently numbered 1…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 K pairs of stalls (1 ≤ K ≤ 100,000). For the ith such pair, you are told two stalls $s_i$ and $t_i$ , 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 $s_i$ to $t_i$, then it counts as being pumped through the endpoint stalls $s_i$ and $t_i$, as well as through every stall along the path between them.

输入格式：

The first line of the input contains N and K.

The next N−1 lines each contain two integers x and y (x≠y) describing a pipe between stalls x and y.

The next K lines each contain two integers s and t 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.

样例数据：

输入 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

输出 9

【分析】

题意就去洛谷上去看啦~~~

其实这道题不用树剖啊那些高级数据结构，只用树上差分+$lca$就行了

由于这是对于点的差分，我们需要在两个端点分别 $+1$，$lca-1$，$lca$ 的父亲 $-1$

举个例子，假设有一条边 $(x,y)$，差分数组为 $d_i$，那么就 $d_x++,d_y++,d_{lca}--,d_{fa[lca]}--$

那么从根开始 $dfs$，按照常规的差分操作，对每个点操作后的 $d_i$ 取个 $max$ 就行了

【代码】

#include<cstdio> #include<cstring> #include<algorithm> #define N 100005 using namespace std; int n,k,t,Max; int first[N],v[N],nxt[N]; int d[N],dep[N],fa[N][20]; void add(int x,int y) { t++; nxt[t]=first[x]; first[x]=t; v[t]=y; } void getfather(int x) { int i,j; for(i=1;i<=17;++i) fa[x][i]=fa[fa[x][i-1]][i-1]; for(i=first[x];i;i=nxt[i]) { j=v[i]; if(j!=fa[x][0]) { fa[j][0]=x; dep[j]=dep[x]+1; getfather(j); } } } int Lca(int x,int y) { int i; if(dep[x]<dep[y]) swap(x,y); for(i=17;~i;--i) if(dep[fa[x][i]]>=dep[y]) x=fa[x][i]; if(x==y) return x; for(i=17;~i;--i) if(fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i]; return fa[x][0]; } void dfs(int x) { int i,j; for(i=first[x];i;i=nxt[i]) { j=v[i]; if(j!=fa[x][0]) { dfs(j); d[x]+=d[j]; } } Max=max(Max,d[x]); } int main() { int x,y,i,j; scanf("%d%d",&n,&k); for(i=1;i<n;++i) { scanf("%d%d",&x,&y); add(x,y),add(y,x); } dep[1]=1,getfather(1); for(i=1;i<=k;++i) { scanf("%d%d",&x,&y); int lca=Lca(x,y); d[x]++,d[y]++,d[lca]--,d[fa[lca][0]]--; } dfs(1); printf("%d",Max); return 0; }