POJ 3469 min cut Dual Core CPU

Test instructions

A dual-core CPU runs n modules, each of which costs AI and bi for each module running on two cores.

At the same time, there is M-to-module data exchange, if the two modules are not running on the same core requires additional cost.

Minimum cost for running N modules.

Analysis:

This is a set partitioning problem, dividing the two modules into two sets, a module in a collection running on kernel A, and one running on the nuclear b.

Add a source point S and a meeting point T, each of which has an edge over the source and sink points, and the capacity is spent running on that core.

Then between two module pairs the capacity is a two-way side for extra cost.

A cut in the figure corresponds to the division of a set, and the minimum cut is the minimum total cost.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <vector>6#include <queue>7 using namespacestd;8 9 Const intMAXN =20000+Ten;Ten Const intINF =0x3f3f3f3f; One  A intN, S, T; - intN, M; -  the structEdge - { -     int  from, to, cap, flow; -Edge (intUintVintCintf): from(U), to (v), Cap (c), Flow (f) {} + }; -  +Vector<edge>edges; Avector<int>G[MAXN]; at  - voidInit () - { - edges.clear (); -      for(inti =0; I < n; i++) g[i].clear (); - } in  - voidAddedge (intUintVintc) to { +Edges.push_back (Edge (U, V, C,0)); -Edges.push_back (Edge (V, U,0,0)); the     intm =edges.size (); *G[u].push_back (M-2); $G[v].push_back (M-1);Panax Notoginseng } -  the BOOLVIS[MAXN]; + intD[MAXN], CUR[MAXN]; A  the BOOLBFS () + { -memset (Vis,false,sizeof(Vis)); $Vis[s] =true; $queue<int>Q; - Q.push (s); -D[s] =0; the  -      while(!q.empty ())Wuyi     { the         intU =Q.front (); Q.pop (); -          for(inti =0; I < g[u].size (); i++) Wu         { -edge& e =Edges[g[u][i]]; About             intv =e.to; $             if(!vis[v] && e.cap >E.flow) -             { -VIS[V] =true; -D[V] = D[u] +1; A Q.push (v); +             } the         } -     } $  the     returnVis[t]; the } the  the intDFS (intUinta) - { in     if(U = = T | | a = =0)returnA; the     intFlow =0, F; the      for(int& i = Cur[u]; I < g[u].size (); i++) About     { theedge& e =Edges[g[u][i]]; the         intv =e.to; the         if(D[v] = = D[u] +1&& (f = DFS (V, Min (A, e.cap-e.flow))) >0) +         { -Flow + =F; theE.flow + =F;BayiA-=F; theedges[g[u][i]^1].flow-=F; the             if(A = =0) Break; -         } -     } the     returnflow; the } the  the intMaxflow () - { the     intFlow =0; the      while(BFS ()) the     {94memset (cur,0,sizeof(cur)); theFlow + =DFS (S, INF); the     } the     returnflow;98 } About  - intMain ()101 {102      while(SCANF ("%d%d", &n, &m) = =2)103     {104n = n +2; the init ();106s =0, T = N +1;107          for(inti =1; I <= N; i++)108         {109             intA, B; scanf"%d%d", &a, &b); the Addedge (S, I, a);111 Addedge (i, T, b); the         }113          while(m--) the         { the             intA, B, w; scanf"%d%d%d", &a, &b, &W); the Addedge (A, B, W);117 Addedge (b, A, w);118         }119  -printf"%d\n", Maxflow ());121     }122 123     return 0;124}

code June

POJ 3469 min cut Dual Core CPU