Poj3436 ACM Computer Factory, maximum stream, output path

Poj3436 ACM Computer Factory, maximum stream, output path
POJ 3436 ACM Computer Factory

Computer companies have N computers, each of which generates Qi in a unit of time.

A computer consists of P parts. Each machine can only work on a semi-finished computer with some parts (or an empty computer with nothing) it turns into a semi-finished computer or a complete computer with other parts (or some parts may be removed ). Ask the computer company for the maximum unit time output, and which machines have a cooperative relationship, that is, the machines on which one machine delivers its products for processing.

Sample input
3 4
15 0 0 0 0 1 0
10 0 0 0 0 1 1
30 0 1 2 1 1 1
3 0 2 1 1 1 1

Sample output
25 2
1 3 15
2 3 10

Input: The computer consists of three parts. There are 4 machines in total. The output of machine 1 is 15. You can add Part 2 to the empty computer, machine 2 can add Part 2 and Part 3 to an empty computer, machine 3 can turn a computer with both Part 2 and Part 3 into a finished product (each part has one)

Output: the maximum output per unit time is 25. Two machines have a cooperative relationship,
Machine 1 needs to process 15 computers to machine 3 within the unit time
Machine 2 needs to process 10 computers to machine 3 within the unit time

The data size is small. You can use EdmondsKarp. The main question is not only the maximum flow value, but also the edge with flow.

Think about it carefully. You don't need to split this question!

#include
 
  #include
  
   #include#include
   
    #include
    
     using namespace std;typedef long long LL;const int maxn = 60;const int inf = 0x7fffffff;struct Edge {    int from,to,cap,flow;    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f) {}};struct EdmondsKarp {    int n,m;    vector
     
       edges;    vector
      
        G[maxn]; int p[maxn]; int a[maxn]; void init(int n) { this->n=n; for(int i=0; i
       
         Q; Q.push(s); a[s]=inf; while(!Q.empty()) { int x=Q.front(); Q.pop(); for(int i=0; i
        
         e.flow) { a[e.to]=min(a[x],e.cap-e.flow); p[e.to]=G[x][i]; Q.push(e.to); if(e.to==t) break; } } if(a[t])break; } if(!a[t])break; for(int u=t; u!=s; u=edges[p[u]].from) { edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; } flow+=a[t]; } return flow; }};EdmondsKarp solver;int P, N;int w[maxn], in[maxn][12], out[maxn][12];int print[maxn][5], cnt;int main(){ int i, j; bool flag; while(~scanf("%d%d", &P, &N)) { int s = 0, t = N+1; solver.init(N+2); for(int i=1; i<=N; ++i) { scanf("%d", &w[i]); flag = true; for(j=0; j
         
          0 && e.to != t && e.from != s) { print[cnt][0] = e.from; print[cnt][1] = e.to; print[cnt++][2] = e.flow; } } printf("%d\n", cnt); for(i=0; i