BZOJ1491 [NOI2007] Social network

Konjac Konjac is the No. 500 AC of the question 233

Yarn, at one glance it looks very high-end, there is a complex formula ...

Later found ... Is... Floyd Water Problem

D[I][J] represents the shortest distance between I, J, Cnt[i][j] represents the shortest distance between I and J

Then use the multiplication principle to update cnt[i][j] can

1 /**************************************************************2 problem:14913 User:rausen4 language:c++5 result:accepted6 time:104 Ms7 memory:936 KB8 ****************************************************************/9  Ten#include <cstdio> One#include <cstring> A   - using namespacestd; -typedefLong Longll; thetypedefDoubleLF; - Const intN = the; -   - intN, M; + intD[n][n]; - ll Cnt[n][n]; + lf ans; A   atInlineintRead () { -     intx =0, SGN =1; -     CharCH =GetChar (); -      while(Ch <'0'||'9'<ch) { -         if(ch = ='-') SGN =-1; -CH =GetChar (); in     } -      while('0'<= CH && Ch <='9') { tox = x *Ten+ CH-'0'; +CH =GetChar (); -     } the     returnSGN *x; * } $  Panax Notoginseng intMain () { -     intI, J, K, X, Y, Z; then = Read (), M =read (); +memset (D,127/3,sizeof(d)); A      for(i =1; I <= m; ++i) { thex = Read (), y = read (), z =read (); +D[x][y] = d[y][x] =Z; -Cnt[x][y] = cnt[y][x] =1; $     } $      for(k =1; K <= N; ++k) -          for(i =1; I <= N; ++i)if(I! =k) -              for(j =1; J <= N; ++J)if(J! = i && J! =k) the                 if(D[i][j] > D[i][k] +D[k][j]) -D[I][J] = D[i][k] +D[k][j],WuyiCNT[I][J] = cnt[i][k] *Cnt[k][j]; the                 Else if(D[i][j] = = D[i][k] +D[k][j]) -CNT[I][J] + = cnt[i][k] *Cnt[k][j]; Wu      for(k =1; K <= N; ++k) { -Ans =0; About          for(i =1; I <= N; ++i)if(I! =k) $              for(j =1; J <= N; ++J)if(J! = i && J! =k) -                 if(D[i][j] = = D[i][k] +D[k][j]) -Ans + = (LF) (cnt[i][k] * cnt[k][j])/Cnt[i][j]; -printf"%.3lf\n", ans); A     } +     return 0; the}

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BZOJ1491 [NOI2007] Social network