Luogu P2886 [USACO07NOV] cattle relay Cow Relays | shortest doubling

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The meaning of problems: given an undirected connected graph, to find S N E via the shortest sides.

Data range: edge number \ (\ Le 100 \) , apex ID \ (\ le1000 \) , \ (N \ LE1 of \ Times. 6 10 ^ \)

answer:

There are three shortest solution, such data may be used range \ (Floyd \)

Can be \ (f [i] [j ] [k] \) represents the \ (I \) to \ (J \) after \ (K \) shortest edges, apparently TLE

Consider doubled. Pretreatment \ (K = 2 ^ K \) , this time \ (f [i] [j ] [k] = min \ {f [i] [l] [k-1] + f [l] [j] [k-1] \} \ )

The \ (n-\) binary split, apparently the first \ (X \) bit \ (1 \) , direct calling \ (F [] [] [X] \) .

Followed by almost the same method can be determined through \ (K \) shortest of the edges.

Note that, according to the vertex run \ (floyd \) time out, so it should be discrete.

The Code

#include<bits/stdc++.h>
#define s(S) 1-S%2
#define g(S) S%2
using namespace std;
const int oo=1000000000;
int n,t,s,e,u,v,c,maxp;
int f[1010][1010][20],ans[1010][1010][2],num[1010],snum;
int main()
{
    cin>>n>>t>>s>>e;
    num[s]=1;num[e]=2;snum=2;s=1;e=2;
    for (int i=1;i<=t;i++)
    {
        cin>>c>>u>>v;
        if (num[u]) u=num[u];else num[u]=++snum,u=num[u];
        if (num[v]) v=num[v];else num[v]=++snum,v=num[v];
        f[u][v][0]=f[v][u][0]=c;
        maxp=max(maxp,max(u,v));
    }   
    for (int k=0;(1<<k)<=n;k++)
    for (int i=1;i<=maxp;i++)
      for (int j=1;j<=maxp;j++)
      {
        if (k!=0||((!f[i][j][k]))) f[i][j][k]=oo;
      }
     for (int k=1;(1<<k)<=n;k++)
      for (int l=1;l<=maxp;l++)
        for (int i=1;i<=maxp;i++)
          for (int j=1;j<=maxp;j++)
          {
              f[i][j][k]=min(f[i][j][k],f[i][l][k-1]+f[l][j][k-1]);
          }//预处理
    int S=0,sum=1,S1=0;
    while (n&&!(n&1))
    {
        n>>=1;S1++;
    }
    for (int i=1;i<=maxp;i++)
        for (int j=1;j<=maxp;j++)
        {
            ans[i][j][S]=f[i][j][S1];
        }
    n>>=1;S1++;
    while (n&&!(n&1))
    {
        n>>=1;S1++;
    }
    for (int i=1;i<=maxp;i++)
        for (int j=1;j<=maxp;j++)
        {
            ans[i][j][1]=oo;
        }
    while (n)
    {
        S++;S%=2;
        for (int l=1;l<=maxp;l++)
        for (int i=1;i<=maxp;i++)
        for (int j=1;j<=maxp;j++)
        {
            ans[i][j][S]=min(ans[i][j][S],ans[i][l][s(S)]+f[l][j][S1]);
        }
        n>>=1;S1++;
        while (n&&!(n&1))
        {
            n>>=1;S1++;
        }
        for (int i=1;i<=maxp;i++)
        for (int j=1;j<=maxp;j++)
        {
            ans[i][j][s(S)]=oo;
        }
    }//求ans，与上面几乎一致
    cout<<ans[s][e][S]<<endl;
    return 0;
}