dag的半连通分量+dp

最新推荐文章于 2025-12-06 19:00:36 发布

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该博客探讨了如何在有向无环图中删除边，使得所有顶点的入度和出度减少或为0，并求解在这种情况下最大可能的「可爱」集合（即在非删除边上的强连通子图）的大小。文章通过拓扑排序和动态规划给出了解决方案，并提供了输入输出示例及代码实现。

G. Remove Directed Edges

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a directed acyclic graph, consisting of nn vertices and mm edges. The vertices are numbered from 11 to nn. There are no multiple edges and self-loops.

Let invinv be the number of incoming edges (indegree) and outvoutv be the number of outgoing edges (outdegree) of vertex vv.

You are asked to remove some edges from the graph. Let the new degrees be in′vin′v and out′vout′v.

You are only allowed to remove the edges if the following conditions hold for every vertex vv:

in′v<invin′v<inv or in′v=inv=0in′v=inv=0;
out′v<outvout′v<outv or out′v=outv=0out′v=outv=0.

Let's call a set of vertices SS cute if for each pair of vertices vv and uu (v≠uv≠u) such that v∈Sv∈S and u∈Su∈S, there exists a path either from vv to uu or from uu to vv over the non-removed edges.

What is the maximum possible size of a cute set SS after you remove some edges from the graph and both indegrees and outdegrees of all vertices either decrease or remain equal to 00?

Input

The first line contains two integers nn and mm (1≤n≤2⋅1051≤n≤2⋅105; 0≤m≤2⋅1050≤m≤2⋅105) — the number of vertices and the number of edges of the graph.

Each of the next mm lines contains two integers vv and uu (1≤v,u≤n1≤v,u≤n; v≠uv≠u) — the description of an edge.

The given edges form a valid directed acyclic graph. There are no multiple edges.

Output

Print a single integer — the maximum possible size of a cute set SS after you remove some edges from the graph and both indegrees and outdegrees of all vertices either decrease or remain equal to 00.

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
const int N=2e5+5;
int h[N],e[N],ne[N],idx;
int din[N],dout[N],d[N];
int q[N];
int dp[N];
int n,m;
void add(int a,int b)
{
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void topsort()
{
    int hh=0,tt=-1;
    for(int i=1;i<=n;i++)
    {
        if(!d[i]) q[++tt]=i;
    }
    while(hh<=tt)
    {
        int now=q[hh++];
        for(int i=h[now];~i;i=ne[i])
        {
            int to=e[i];
            if(dout[now]>=2&&din[to]>=2) dp[to]=max(dp[to],dp[now]+1);
            if(--d[to]==0) q[++tt]=to;
        }
    }
}
int main()
{
    cin>>n>>m;
    memset(h,-1,sizeof h);
    for(int i=1;i<=n;i++) dp[i]=1;
    for(int i=1;i<=m;i++)
    {
        int a,b;
        cin>>a>>b;
        add(a,b);
        din[b]++,dout[a]++,d[b]++;
    }
    topsort();
    int res=0;
    for(int i=1;i<=n;i++) res=max(res,dp[i]);
    cout<<res<<endl;
}