LS 22 Longest path on DAG(最短路+SPFA）

最新推荐文章于 2021-08-23 21:00:25 发布

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本文探讨了在有向无环图中寻找最长路径的问题，通过构建额外的源点和汇点，利用SPFA算法求解源点到汇点的最长路径，从而得到图中最长路径及其顺序。

Longest path on DAG

Find the longest path on directed acyclic graph G.

Input

The first line contains two integers n, m, which denote the number of vertices and edges.

The following m lines contains two integer ai, bi, which denote edge ai→bi.

(1≤n≤105,1≤m≤106,1≤ai<bi≤n)

Ouptut

The first line contains an integer l, which denotes the length of longest path.

The second line contains l+1 integers, which denote the longest path with smallest lexicographic order.

Sample input

Sample output

2
1 2 4

思路：建边，建个源点0和汇点，然后依次在源点和汇点与中间所有点连边。求源点和汇点的最长路就是答案。

本题需要从汇点往源点做最短路

#include<iostream>
#include<cstring>
using namespace std;
const int mm=6e6+9;
int ver[mm],next[mm],head[mm],edge,q[mm],dis[mm],fa[mm];
bool vis[mm];
int n,m;
void prepare()
{
  for(int i=0;i<n+2;i++)head[i]=-1,fa[i]=-1;edge=0;
}
void add(int aa,int bb)
{
  ver[edge]=bb;next[edge]=head[aa];head[aa]=edge++;
}
int spfa()
{ memset(dis,-1,sizeof(dis));
  memset(vis,0,sizeof(vis));
  int l=0,r=1,u,v;
  q[l]=0;vis[0]=1;dis[0]=0;
  while(l^r)
  {u=q[l++];l%=mm;vis[u]=0;
    for(int i=head[u];i!=-1;i=next[i])
    { v=ver[i];
      if(dis[v]<dis[u]+1)
      {
        dis[v]=dis[u]+1;fa[v]=u;
        if(!vis[v])
        q[r++]=v,r%=mm,vis[v]=1;
      }
      else if(dis[v]==dis[u]+1&&fa[v]>u)fa[v]=u;
    }
  }
  cout<<dis[n+1]-2<<"\n";
  int pos=0,zz=fa[n+1];
  while(zz!=-1)
  {
    ver[pos++]=zz;
    zz=fa[zz];
  }
  for(int i=0;i<pos-2;i++)
    cout<<ver[i]<<" ";
    cout<<ver[pos-2]<<"\n";
}
int main()
{
  while(cin>>n>>m)
  { int a,b;
    prepare();
    for(int i=0;i<m;i++)
    {
      cin>>a>>b;add(b,a);///建反向边
    }
    for(int i=1;i<=n;i++)
      add(0,i),add(i,n+1);
    spfa();
  }
}