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[Luogu 3128] USACO15DEC Max Flow

<题目链接>

最近跟 LCA 干上了…

树剖好啊，我再也不想写倍增了。

以及似乎成功转成了空格选手 qwq。

对于每两个点 S and T，求一下 LCA 顺便树上差分，最后求差分数组的前缀和并找出最大值输出就行了。

（PS：最近考前训练不开 C++11，所以如果看见我写了奇怪的 define 请自动无视QAQ！）

#include <algorithm>
#include <cstdio>

#define nullptr NULL

const int MAXN = 50010; 

int n, m;

namespace HLD
{
    int num, qwq[MAXN]; 
    struct Node
    {
        int depth, father, son, top, size, DFN; 
    }s[MAXN]; 
    class Graph
    {
        private: 
            struct Edge
            {
                int to; 
                Edge *next; 
                Edge(int to, Edge* next): to(to), next(next) {}
                ~Edge(void)
                {
                    if(next != nullptr)
                        delete next; 
                }
            }*head[MAXN]; 
            void AddEdges(int u, int v)
            {
                head[u] = new Edge(v, head[u]); 
                head[v] = new Edge(u, head[v]); 
            }
            void DFS1(int u, int k)
            {
                s[u].depth = k; 
                s[u].size = 1; 
                int v; 
                for(Edge *i = head[u]; i != nullptr; i = i -> next)
                    if(!s[v = i -> to].size)
                    {
                        DFS1(v, k + 1); 
                        s[u].size += s[v].size; 
                        s[v].father = u; 
                        if(s[v].size > s[s[u].son].size)
                            s[u].son = v; 
                    }
            }
            void DFS2(int u, int top)
            {
                s[u].top = top; 
                s[u].DFN = ++num; 
                if(s[u].son)
                    DFS2(s[u].son, top); 
                int v; 
                for(Edge *i = head[u]; i != nullptr; i = i -> next)
                    if(!s[v = i -> to].top)
                        DFS2(v, v); 
            }
            void Modify(int x, int y)
            {
                ++qwq[s[x].DFN]; 
                --qwq[s[y].DFN + 1]; 
            }
        public: 
            Graph(int n)
            {
                std :: fill(head + 1, head + n + 1, (Edge*)nullptr); 
                for(int i = 1, x, y; i < n; ++i)
                {
                    scanf("%d %d", &x, &y); 
                    AddEdges(x, y); 
                }
                DFS1(1, 1); 
                DFS2(1, 1); 
            }
            ~Graph(void)
            {
                for(int i = 1; i <= n; ++i)
                    delete head[i]; 
            }
            void Run(int x, int y)
            {
                int a, b; 
                while((a = s[x].top) ^ (b = s[y].top))
                    if(s[a].depth > s[b].depth)
                    {
                        Modify(a, x); 
                        x = s[a].father; 
                    }
                    else
                    {
                        Modify(b, y); 
                        y = s[b].father; 
                    }
                if(s[x].depth < s[y].depth)
                    Modify(x, y); 
                else
                    Modify(y, x); 
            }
            int Answer(void)
            {
                int ans = qwq[1]; 
                for(int i = 2; i <= n; ++i)
                    ans = std :: max(ans, qwq[i] += qwq[i-1]); 
                return ans; 
            }
    }*G; 
}

int main(void)
{
    scanf("%d %d", &n, &m); 
    HLD :: G = new HLD :: Graph(n); 
    for(int i = 1, x, y; i <= m; ++i)
    {
        scanf("%d %d", &x, &y); 
        HLD :: G -> Run(x, y); 
    }
    printf("%d\n", HLD :: G -> Answer()); 
    return 0; 
}

谢谢阅读。

[Luogu 3128] USACO15DEC Max Flow

标签：题目   clu   \n   top   public   lca   flow   return   amp   

原文地址：https://www.cnblogs.com/Capella/p/9892486.html