UVA 820---POJ 1273 Max stream

Looking for a long time the difference between the two ... UVA820 wa a lot of times. However, after the template, you can find the maximum flow between any two points.

UVA is a graph, so there may be heavy edges, POJ 1273 is a graph, and is the single source point for the maximum flow, so when changing the template to pay attention to.

And I actually made a more stupid mistake, in order to re-edge the need to choose the largest, positive solution should be cumulative ....

1#include <stdio.h>2#include <queue>3#include <string.h>4 #defineINF 9999995 using namespacestd;6 ints,t,n,m;7 intmap[407][407],flow[407][407];8 intp[407],a[407];9 intEK (intSintt)Ten { Onequeue<int>Q; A     intsum=0; -memset (Flow,0,sizeof(flow)); -      while(1) the     { -Memset (A,0,sizeof(a)); -a[s]=INF; - Q.push (s); +          while(!q.empty ()) -         { +             intu=Q.front (); A Q.pop (); at              for(intI=1; i<=m; i++) -             { -                 if(!a[i]&&flow[u][i]<Map[u][i]) -                 { -p[i]=u; - Q.push (i); ina[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-Flow[u][i]; -                 } to             } +         } -         if(!a[t]) Break; the          for(inti=t; I!=s; I=P[i]) *         { $flow[p[i]][i]+=A[t];Panax Notoginsengflow[i][p[i]]-=A[t]; -         } thesum+=A[t]; +     } A     returnsum; the } + intMain () - { $     intcas=1; $      while(SCANF ("%d",&m), m) -     { -scanf"%d%d%d",&s,&t,&n); thememset (Map,0,sizeof(map)); -         inta,b,c;Wuyi          for(intI=0; i<n; i++) the         { -scanf"%d%d%d",&a,&b,&c); Wumap[a][b]+=C; -map[b][a]=Map[a][b]; About         } $         intmaxx=EK (s,t); -printf"Network%d\n", cas++); -printf"The bandwidth is%d.\n\n", Maxx); -     } A     return 0; +}

UVA 820 Direct ek template on

Can be compared with the POJ 1273 map

commented out a plus edge

POJ 1273 Examples to

2 2

1 2 3

1 2 5 The answer is 8.

2 2

1 2 3

2 1 5 The answer is 3.

UVA 820---POJ 1273 Max stream