POJ 2135 Farm Tour "minimum cost maximum Flow"

The topic of the first fee flow---

In fact---still do not understand, only know along the shortest path to find the augmented road

Building Map

Source point to 1 with a capacity of 2 (because to go back and forth), the cost is 0 of the edge

N to a meeting point with a capacity of 2 and a cost of 0 sides

The other one is the one that was typed in the title.

In addition, the problem is the non-direction side-----

MAXN to 1000, re----

Save a template First------------------------------------------

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <iostream>5#include <algorithm>6#include <Set>7#include <map>8#include <stack>9#include <vector>Ten#include <queue> One#include <string> A using namespacestd; -  -typedefLong LongLL; the Const intMAXN = the; - Const intINF = (1<< -) -1; -   - intfirst[maxn],vis[maxn],dis[maxn],pos[maxn],ecnt,size; +  - structedge{ +     intV,next,cap,cost; A} e[Ten*MAXN]; at  - voidinit () { -ECNT =0; -memset (first,-1,sizeof(first)); - } -    in voidAdd_edge (intUintVintCapintCost ) { -E[ECNT].V =v; toE[ecnt].cap =cap; +E[ecnt].cost =Cost ; -E[ecnt].next =First[u]; theFirst[u] = ecnt++; *      $E[ECNT].V =u;Panax NotoginsengE[ecnt].cap =0; -E[ecnt].cost =-Cost ; theE[ecnt].next =First[v]; +FIRST[V] = ecnt++; A } the    + BOOLSPFA (intSintt) - { $     intu,v,i; $Queue <int>Q; -memset (Vis,0,sizeof(Vis)); -      for(i=0; I <= size;i++) dis[i]=INF; the      -dis[s]=0;Wuyivis[s]=1; the Q.push (s); -      Wu      while(!Q.empty ()) { -U=q.front (); Q.pop (); vis[u]=0; About          for(i = first[u]; ~i;i =E[i].next) { $v=e[i].v; -                if(E[i].cap >0&& Dis[u]+e[i].cost <Dis[v]) { -dis[v]=dis[u]+E[i].cost; -pos[v]=i; A                   if(!Vis[v]) { +vis[v]=1; the Q.push (v); -                   } $                } the           } the     } the     returnDIS[T]! =INF; the } -  inLL MCMF (intSintt) the { the     inti; AboutLL cost=0, flow=0; the      while(SPFA (s,t)) { the         intD=INF; the          for(i = t;i! = S;i = e[pos[i]^1].v) { +D =min (d,e[pos[i]].cap); -         } the          for(i = t;i! = S;i = e[pos[i]^1].v) {BayiE[pos[i]].cap-=D; thee[pos[i]^1].cap + =D; the         } -Flow + =D; -Cost + = dis[t]*D; the     } the     returnCost ; the } the  -  the intMain () { the     intn,m; the      while(SCANF ("%d%d", &n,&m)! =EOF) {94 init (); theSize = n+1; theAdd_edge (0,1,2,0); Add_edge (1,0,2,0); theAdd_edge (n+1N2,0); Add_edge (n,n+1,2,0);98          About          for(inti =0; I < m;i++){ -             intu,v,w;101scanf" %d%d%d",&u,&v,&W);102Add_edge (U,v,1, W);103Add_edge (V,u,1, W);104         } theprintf"%i64d\n", MCMF (0, n+1));106     }107     return 0;108}

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POJ 2135 Farm Tour "minimum cost maximum Flow"