[Code+#4]最短路

[\(Code\)+#4]最短路

链接:https://www.luogu.com.cn/problem/P4366

题面:给定一个\(n\)个点，\(m\)条边的无向图，若任意点对\((i,j)\)之间还有一条长为\((i xor j)\times C\)的边，求\(A\)到\(B\)的最短路。

题解:\((i xor j)\times C\)这个式子看似很毒瘤，实际上我们可以把每一位拆开来看，若要从\(9\)走到\(7\)：

\(((1001)_{2} xor (111)_{2})\times C\)

\(=(1000)_{2}\times C+(10)_{2}\times C+(100)_{2}\times C\)

\(=((1001)_{2}xor(1)_{2})\times C+((1)_{2}xor(11)_{2})\times C+((11)_{2}xor(111)_{2})\times C\)

上述等式的含义即为\((1001)_{2}->(111)_{2}\)等价于\((1001)_{2}->(1)_{2}->(11)_{2}->(111)_{2}\)

推广即得,任意\((i,j)\)的边，都可以被拆成若干个\((i,(i xor 2^t))\)的边的权值和，每一个点\(i\)向\((i xor 2^t)\)连边即可。

#include<iostream>
#include<queue>
using namespace std;
struct node
{
  int v,data,nxt;
};
struct reads
{
  int num,data;
  bool operator < (const reads &a)const
  {
    return data>a.data;
  }
};
reads tmp;
priority_queue<reads>q;
node edge[4000001];
int head[200001],len,dis[200001],n,m,C;
bool used[200001];
void add(int x,int y,int z)
{
  edge[++len].v=y;
  edge[len].data=z;
  edge[len].nxt=head[x];
  head[x]=len;
  return;
}
reads make_reads(int x,int y)
{
  tmp.num=x;
  tmp.data=y;
  return tmp;
}
void dijkstra(int x)
{
  q.push(make_reads(x,0));
  for (int i=0;i<=n;++i)
    dis[i]=(i!=x)*1e9;
  int top;
  while (!q.empty())
    {
      top=q.top().num;
      q.pop();
      if (used[top])
	continue;
      used[top]=1;
      for (int i=head[top];i>0;i=edge[i].nxt)
	if (dis[edge[i].v]>dis[top]+edge[i].data)
	  {
	    dis[edge[i].v]=dis[top]+edge[i].data;
	    q.push(make_reads(edge[i].v,dis[edge[i].v]));
	  }
    }
}
int main()
{
  int x,y,z,A,B;
  cin>>n>>m>>C;
  for (int i=1;i<=m;++i)
    {
      cin>>x>>y>>z;
      add(x,y,z);
    }
  for (int i=0;i<=n;++i)
    for (int k=1;k<=2*n;k*=2)
      if ((i^k)<=n)
	add(i,i^k,k*C);
  cin>>A>>B;
  dijkstra(A);
  cout<<dis[B]<<endl;
  return 0;
}