USACO December 铂金Maxflow

USACO 2015 December Contest, Platinum

Problem 1. Max Flow

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 i th such pair, you are told two stalls si and ti , 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 si to ti , then it counts as being pumped through the endpoint stalls si and ti , as well as through every stall along the path between them.

INPUT FORMAT (file maxflow.in):

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.

OUTPUT FORMAT (file maxflow.out):

An integer specifying the maximum amount of milk pumped through any stall in the barn.

SAMPLE INPUT:

SAMPLE OUTPUT:

Problem credits: Brian Dean

译：

农夫约翰已经安装了N-1个管道系统给N（2≤N≤50,000）个摊位输送牛奶，方便编号为1... N。每个管道连接一对摊位，所有的摊位都通过管道路径连接起来。

FJ在K对点（1≤K≤100,000）之间抽取牛奶。对于第i个这样的对，你被告知两个点Si和Ti，表示抽取牛奶路径的两个端点，这一路径上都是同一个单位的费率。 FJ关注的是，一些点最终可能会被所有抽取的牛奶通过。请帮助他确定所有点可能通过的最多的牛奶量。如果牛奶正在沿着路径Si到Ti，则认为它将通过端点Si和Ti，以及通过沿它们之间的路径中的每个点。

INPUT FORMAT (file maxflow.in):

The first line of the input contains N and K.

第一行两个整数N和K

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

接下来N-1行，每行两个整数X和Y，描述被一个管子连接的两个点X和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.

接下来K行，每行两个整数 S和T，描述正在被抽取牛奶所在的两个端点

OUTPUT FORMAT (file maxflow.out):

An integer specifying the maximum amount of milk pumped through any stall in the barn.

输出一个整数，表示通过任意点所能抽取的最大牛奶数量。

树上查分

拿到这题第一反应显然是树剖。。但实际上有更加方便的做法，那就是树上差分。

首先我们联想一下对数列进行差分的做法。序列差分这个技巧一般适用于：执行若干次区间加减，到最后再统计每个点的权值。设T是A的差分序列，我们把将a~b这个区间中的每个点加上c的操作（A[i]+=c a<=i<=b）转变成对T的操作T[a]+=c，T[b+1]-=c，最后A[k]的值就是Sum{T[i]}(1<=i<=k)即T的一个前缀和。这个很容易能理解，随便在脑中yy一下或者画画图都能理解。

然后我们再来看下这道题。这道题与序列上有一定相似之处：执行若干次路径上的加减，最后统计每个点的权值。我们可以将在序列上进行差分的方法拓展到树上。对于s~t这个路径上每个点加上c的操作，我们可以对树A的“差分树”T进行操作：T[s]+=c，T[t]+=c，T[LCA(s,t)]-=c，T[Father[LCA(s,t)]]-=c。即给s这个点加上c，给t这个店加上c，给s和t的最近公共祖先和这个祖先的父亲减去c。最后每个点的权值就是这个点的子树权值之和。

这样做为什么是正确的呢？我们来yy一下一般情况：对于一棵树，从s走到t可以这么走：先从s走到LCA(s,t),再从t走到LCA(s,t)。为了更方便接下来的叙述，我们可以想象有两个人分别从s和t走到LCA(s,t)。然后，我们要将路径s~t上每个点加上c，因为我们最后统计的是每个点的子树权值之和，所以说白了就是这个标记要向它的父亲上传；所以当我们给s和t都加上c后，上传到LCA(s,t)的时候就变成了2c，显然在这里是要减去c的；而对于LCA(s,t)的父亲，由于LCA(s,t)的标记也要上传，所以在LCA(s,t)的父亲处再减去一个c。这段话比较长，但是只要深刻理解数列差分的技巧，理解树上差分也是很容易的。

最后说几个实现上的细节。

第一，对于LCA(s,t)，如果它是根，那么就没有将它父亲减去c的操作了。

第二，最后子树统计的时候的方法比较多。可以用类似拓扑排序的思想，从最后一层结点“逐层上传”。也可以用树型dp。这个随个人喜好而定。

具体看代码~

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdlib>

#define maxn 50010

#define maxk 100010

using namespace std;

int n,k,s,t,tot,lca,root=,ans=,f[maxn],deep[maxn],value[maxn],head[maxn],p[maxn][];

struct node{

    int go,next;

}e[maxk];

inline int read(){

    int x=,f=;char ch=getchar();

    while (ch>'' || ch<''){if (ch=='-') f=-;ch=getchar();}

    while (ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void addedge(int &s,int &t){

    e[++tot]=(node){t,head[s]};head[s]=tot;

    e[++tot]=(node){s,head[t]};head[t]=tot;

}

void make_tree(int u,int fa){

    int v;

    for (int i=head[u];i;i=e[i].next){

        v=e[i].go;

        if (v!=fa){

            deep[v]=deep[u]+;

            f[v]=u;

            make_tree(v,u);

        }

    }

}

void prepare(){

    memset(p,-,sizeof(p));

    for (int i=;i<=n;i++) p[i][]=f[i];

    for(int j=;(<<j)<=n;j++)

        for (int i=;i<=n;i++)

            if (p[i][j-]!=-) p[i][j]=p[p[i][j-]][j-];

}

inline int LCA(int a,int b){

    if (deep[a]<deep[b]) swap(a,b);

    int t=trunc(log2(deep[a]));

    for (int i=t;i>=;i--) if (deep[a]-(<<i)>=deep[b]) a=p[a][i];

    if (a==b) return a;

    for (int i=t;i>=;i--)

        if (p[a][i]!=- && p[a][i]!=p[b][i]) a=p[a][i],b=p[b][i];

    return p[a][];

}

int ask(int x){

    for (int i=head[x];i;i=e[i].next){

        if (f[x]!=e[i].go){

            ask(e[i].go);

            value[x]+=value[e[i].go];

        }

    }

    ans=max(ans,value[x]);

}

int main(){

//    freopen("maxflow.in","r",stdin);

//    freopen("maxflow.out","w",stdout);

    n=read();k=read();

    for (int i=;i<n;i++){

        s=read();t=read();

        addedge(s,t);

    }

    deep[root]=;

    make_tree(root,f[root]=);

    prepare();

    for (int i=;i<=k;i++){

        s=read();t=read();

        lca=LCA(s,t);

        value[s]++;value[t]++;value[lca]--;value[f[lca]]--;

    }

    ask(root);

    printf("%d\n",ans);

}