POJ 2112 floyd+dinic Maximum flow + binary minimum

The main topic is:

K-Table milking machine, C cattle, K not more than 30,c not more than 200, each milking machine can work for M-cow, give these cows and machines, between the cow and the machine, the distance between the machines and machines, in order to ensure that the most cows have a machine milking the case, give the distance of the longest one of the cattle moving distance of the minimum value.

First use Floyd to find the shortest distance between any two points, and then use the dichotomy to limit the most moving distance d, when the maximum flow, the search for the augmented path at the same time also judge the distance is not more than D on the line.

1#include <cstdio>2#include <cstring>3#include <queue>4 #define_CLR (x, y) memset (x, y, sizeof (x))5 #defineMin (x, y) (x < y x:y)6 #defineINF 0x3f3f3f3f7 #defineN 10058 using namespacestd;9 Ten intFlow[n][n], dist[n][n]; One intLevel[n]; A intK, C, M, S, T, ln; -  - voidFloyd () the { -      for(intk=1; k<=ln; k++) -          for(intI=1;i<=ln; i++)if(dist[i][k]<INF) -              for(intj=1; j<=ln; J + +) +                 if(dist[i][k]+dist[k][j]<Dist[i][j]) -DIST[I][J] = dist[i][k]+Dist[k][j]; + } A  at BOOLBFS () - { -_CLR (Level,-1); -Level[s] =0; -queue<int>Q; - Q.push (S); in      while(!q.empty ()) -     { to         intU =Q.front (); + Q.pop (); -          for(intI=0; i<=t; i++) the         { *             if(Flow[u][i] && level[i]<0) $             {Panax NotoginsengLevel[i] = Level[u] +1; - Q.push (i); the             } +         } A     } the     returnLevel[t]>0?1:0; + } -  $ intDfsintXintf) $ { -     intA; -     if(x==t)returnF; the      for(intI=0; i<=t; i++) -     {Wuyi         if(Flow[x][i] && level[i]==level[x]+1&& (a=Dfs (i,min (f,flow[x][i )))) the         { -Flow[x][i]-=A; WuFLOW[I][X] + =A; -             returnA; About         } $     } -LEVEL[X] =-1; -     return 0; - } A  +__int64 Dinic (intlen) the {        -     //build a residual network $_CLR (Flow,0); the      for(intI=1; i<=k; i++) theFlow[s][i] =M; the      for(inti=k+1; i<=ln; i++) theFLOW[I][T] =1; -      for(intI=1; i<=k; i++)//Machine in          for(inti=h+1; j<=ln; J + +)//cows theFLOW[I][J] = (dist[i][j]<=len); the  About     //solving the maximum flow the__int64 ans=0, a=0; the      while(BFS ()) the          while(A=dfs (0, INF)) ans + =A; +     returnans; - } the Bayi //The minimum solution satisfying the condition by two-part solution the intSlove () the { -     intL=1, r=100000; -      while(l<=R) the     { the         intMid = (l+r) >>1; the         if(Dinic (mid) >=c) R = mid-1; the         ElseL = mid+1; -     } the     returnl; the } the intMain ()94 { the      while(~SCANF ("%d%d%d", &k, &c, &M)) the     { theln = k +C;98T = ln +1; About_CLR (Dist,0); -          for(intI=1; i<=ln; i++)101              for(intj=1; j<=ln; J + +)102             {103scanf"%d", dist[i]+j);104                 if(dist[i][j]==0) dist[i][j]=INF; the             }106         ////Find the shortest distances between entities107 Floyd ();108 109printf"%d\n", Slove ()); the     }111     return 0; the}

POJ 2112 floyd+dinic Maximum flow + binary minimum