[USACO15DEC]最大流Max Flow

 

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 ($2\le N\le 50,000$), conveniently numbered $1\dots N$. 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 ($1\le K\le 100,000$). For the $i$th such pair, you are told two stalls s_is​i​​ and t_it​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 $K$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_is​i​​ to t_it​i​​, then it counts as being pumped through the endpoint stalls s_is​i​​ and

t_it​i​​, 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 $N$ and $K$.

The next $N-1$ lines each contain two integers $x$ and $y$ ($x\ne y$) 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.

输入输出样例

输入样例#1：

5 103 41 54 25 45 45 43 54 34 31 33 55 41 53 4

输出样例#1：

9

    一道树上差分的裸题，也是因为要做交通运输，所以来看看树上差分的知识，于是便找上了这题，题目描述很简单，就是给你一棵树以及几个操作，这些操作包含两个点，具体内容就是让你把这两个点之间的路径上所有的点点权加一，最后输出点权最大值。

    既然是求树上的两点之间路径，很容易想到LCA，tarjan（N+K）的复杂度足以胜任，接下来便是玄学的推导，到最后你会发现，其实对于读入两个点{u,v},我们要进行的只是tree[u]++,tree[v]++,tree[lca(u,v)]--,tree[father[lca(u,v)]]--;其中tree[]表示每个点的点权，到最后统计的时候再加上这个点所有子节点的点权即是它的最终点权

贴上一段代码，看不懂的地方在评论区里戳我。。

#include#include#include#define N 50010#define K 100010using namespace std;struct node1{int v;};struct node2{int v,id;};vector  pic[N];vector  edge[N];int vis[N],f[N],up[N],ans[K],tree[N],n,k,ANS;int find(int x){if (f[x]==x) return x;return f[x]=find(f[x]);}void tarjan(int u){int i;vis[u]=1;f[u]=u;for (i=0;iANS) ANS=tree[u];}int main(){int i;scanf("%d%d",&n,&k);for (i=1;i