Poj3469 Dual Core CPU --- minimum cut

Poj3469 Dual Core CPU --- minimum cut

A cpu has two cores. To put n modules on one core, the cost of putting them on different cores is given.

In addition, the m pair module is provided. If it is not placed in the same module, additional costs will be incurred. Minimum cost.

For each module, you can select core 1, Core 2, and connected modules.

Based on this, the edge is built, the core 1 is the source point, the Core 2 is the sink point, and the two-way edge is built between the connected modules, and the edge weight is the cost. Find the minimum cut.

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include #include 
      
       #include 
       
        #include 
        #define inf 0x3f3f3f3f#define eps 1e-6#define ll __int64const int maxn=20010;using namespace std;struct node{ int from,to,cap,flow;};struct dinic{ int n,m,s,t; vector
         
           e; vector
          
            g[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; void init(int n) { e.clear(); for(int i=0;i<=n+2;i++) g[i].clear(); } void addedge(int a,int b,int c,int d) { e.push_back((node){a,b,c,0}); e.push_back((node){b,a,d,0}); m=e.size(); g[a].push_back(m-2); g[b].push_back(m-1); } bool bfs() { memset(vis,0,sizeof vis); queue
           
             q; q.push(s); d[s]=0; vis[s]=1; while(!q.empty()) { int x=q.front();q.pop(); for(int i=0;i
            
             ee.flow) { vis[ee.to]=1; d[ee.to]=d[x]+1; q.push(ee.to); } } } return vis[t]; } int dfs(int x,int a) { if(x==t||a==0) return a; int flow=0,f; for(int& i=cur[x];i
             
              0) { ee.flow+=f; e[g[x][i]^1].flow-=f; flow+=f; a-=f; if(a==0) break; } } return flow; } int maxflow(int s,int t) { this->s=s; this->t=t; int flow=0; while(bfs()) { memset(cur,0,sizeof cur); flow+=dfs(s,inf); } return flow; }};dinic solve;int main(){ int i,a,b,cc,m,n,s,t; while(~scanf("%d%d",&n,&m)) { s=0,t=n+1; solve.init(n); for(i=1;i<=n;i++) { scanf("%d%d",&a,&b); solve.addedge(s,i,a,0); solve.addedge(i,t,b,0); } while(m--) { scanf("%d%d%d",&a,&b,&cc); solve.addedge(a,b,cc,cc); } printf("%d\n",solve.maxflow(s,t)); } return 0;}