"Floyd" BZOJ1491: [NOI2007] Social network

Description Solution

N<=100 Natural Association Floyd

Set two arrays d[n][n] minimum distance, t[n][n] minimum number of paths

Update d on the way to update T, multiplication principle

1 if (d[i][j]>d[i][k]+d[k][j]) {2     d[i][j]=d[i][k]+d[k][j]; 3     t[i][j]=t[i][k]*T[k][j]; 4 }5Elseif(d[i][j]==d[i][k]+d[k][j])6     t[i ][J]+=T[I][K]*T[K][J];

The answer to the statistics

1 if (d[i][j]==d[i][k]+d[k][j]) 2     c[k]+= (double) (T[i][k]*t[k][j])/t[i][j];

Feel Floyd can do these things.

Code

1#include <cstdio>2#include <algorithm>3#include <cstring>4 #definell Long Long5 Const intmaxn= the;6 7 ll D[MAXN][MAXN],T[MAXN][MAXN];8 intn,m;9 DoubleC[MAXN];Ten  One intMain () { Ascanf"%d%d",&n,&m); -     intu,v,w; -memset (D,127/3,sizeof(d)); the      for(intI=1; i<=n;i++) -d[i][i]=0, t[i][i]=1; -          -      for(intI=1; i<=m;i++){ +scanf"%d%d%d",&u,&v,&W); -         if(w<D[u][v]) { +d[u][v]=d[v][u]=W; At[u][v]=t[v][u]=1; at         } -         Else if(w==D[u][v]) { -t[u][v]++; -t[v][u]++; -         } -     } in      -      for(intk=1; k<=n;k++) to          for(intI=1; i<=n;i++) +              for(intj=1; j<=n;j++)if(i!=j&&i!=k&&j!=k) { -                 if(d[i][j]>d[i][k]+D[k][j]) { thed[i][j]=d[i][k]+D[k][j]; *t[i][j]=t[i][k]*T[k][j]; $                 }Panax Notoginseng                 Else if(d[i][j]==d[i][k]+D[k][j]) -t[i][j]+=t[i][k]*T[k][j]; the             } +      AMemset (c,0,sizeof(c)); the      for(intI=1; i<=n;i++) +          for(intj=1; j<=n;j++) -              for(intk=1; k<=n;k++)if(i!=j&&i!=k&&j!=k) { $                 if(D[i][j]!=d[i][k]+d[k][j])Continue; $c[k]+= (Double) t[i][k]*t[k][j]/T[i][j]; -             } -      the      for(intI=1; i<=n;i++) -printf"%.3lf\n", C[i]);Wuyi     return 0; the}

"Floyd" BZOJ1491: [NOI2007] Social network