POJ 1698 (multi-match for binary graphs)

Reprint: Http://www.cppblog.com/MatoNo1/archive/2011/03/26/142766.aspx

　　We know that in a graph, the maximum number of matches for a single edge of each point is the largest matching problem of a binary graph. However, there is a situation where each point can match multiple edges, but there is an upper limit, which is assumed to be L. That is, the maximum point I can be associated with the Li Bar.

Binary graph multiple maximum match:

1. Create a source point S and a meeting point T.

The 2.S points to the X vertex, and the capacity is the L value of the point within X. The y vertex points to T, and the capacity is the L value of the point within Y.

3. Each side of the original image still exists in the new diagram, with a capacity of 1.

Then the maximum flow of S to T is the multiple maximum match.

For example POJ1698:

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 #defineMax (x, y) (x > Y x:y)7 #defineINF 0x3f3f3f3f8 #defineN 4009 using namespacestd;Ten  One intEdge[n][n], dist[n]; A intT, S, Sum, N; - BOOLUsed[n]; -  the BOOLBFS () - { -_CLR (Dist,-1); -queue<int>Q; +Dist[s] =0; - Q.push (S); +      while(!q.empty ()) A     { at         intU =Q.front (); - Q.pop (); -          for(intv=0; v<=t; v++) -         { -             if(Edge[u][v] && dist[v]<0) -             { inDIST[V] = Dist[u] +1; - Q.push (v); to             } +         } -     } the     returnDist[t]>0?1:0; * } $ Panax Notoginseng intDfsintUintAlpha) - { the     intA; +     if(u==t)returnAlpha; A      for(intI=0; i<=t; i++) the     { +         if(Edge[u][i] && dist[i]==dist[u]+1&& (a=DFS (i, Min (Alpha, Edge[u][i] )))) -         { $Edge[u][i]-=A; $Edge[i][u] + =A; -             returnA; -         } the     } -Dist[u] =-1;Wuyi     return 0; the } - voiddinic () Wu { -     intans=0, a=0; About      while(BFS ()) $          while(A=dfs (0, INF)) ans + =A; -printf ("%s\n", Ans==sum?"Yes":"No"); - } -  A intMain () + { the     intK, week[Ten]; -scanf"%d", &K); $      while(k--) the     { the         intDay =0, D, W; thescanf"%d", &n); theSum =0, s=0; -_CLR (Week,0); in_CLR (Edge,0); the          for(intI=1; i<=n; i++)//1--n between the movies, N+1---t represents the number of days! the         { About              for(inti1=0; i1<7; i1++) scanf ("%d", week+i1); thescanf"%d%d", &d, &W); theDay =Max (Day, W); theEdge[s][i] = D;//The source points to each festival point x to connect a weight to the number of days required to take the movie. +Sum + =D; -  the              for(intj=0; j<w; J + +)//within a W week to complete.Bayi             { the                  for(intk=0; k<7; k++) the                     if(Week[k]) -edge[i][n+j*7+k+1] =1;//Part I movies can be taken at week K within W week. -             } theT = day*7+n+1; the              for(inti=n+1; i<t; i++) theEDGE[I][T] =1; the         } - dinic (); the     } the     return 0; the}

POJ 1698 (multi-match for binary graphs)