Network Flow: Max-Flow, Min-Cut

9. Practice Problems

1. Maximum bipartite matching: 5 students applying to 5 universities. Find max assignment.
   ৫ জন ছাত্র ৫টি বিশ্ববিদ্যালয়ে আবেদন করেছে। সর্বাধিক বরাদ্দ বের করুন।
   ✨ Show Answer (উত্তর দেখুন)

   ans1.cpp

   #include <bits/stdc++.h>
   using namespace std;
   int L=5, R=5;
   vector<vector<int>> adj;
   vector<int> matchR;
   vector<bool> vis;
   bool aug(int u){for(int v:adj[u]){if(vis[v])continue;vis[v]=1;if(matchR[v]==-1||aug(matchR[v])){matchR[v]=u;return 1;}}return 0;}
   int main(){
       adj.assign(L,{});
       adj[0]={0,1}; adj[1]={1,2}; adj[2]={0,3}; adj[3]={2,4}; adj[4]={3,4};
       matchR.assign(R,-1); int tot=0;
       for(int u=0;u<L;++u){vis.assign(R,0);if(aug(u))++tot;}
       cout<<"Matching = "<<tot<<"\n";
   }

   Run Copy

2. Minimum vertex cover in a bipartite graph via König's theorem (= max matching size).
   König's theorem অনুযায়ী bipartite graph-এ minimum vertex cover = max matching। প্রমাণ করুন।
   ✨ Show Answer (উত্তর দেখুন)

   Answer: König's theorem states that in any bipartite graph, the size of a minimum vertex cover equals the size of a maximum matching. Construction: after computing the matching, let U = unmatched left nodes. Run an alternating BFS from U (alternating non-matching/matching edges). Let Z = visited nodes. Then the cover is (Left \ Z) ∪ (Right ∩ Z); its size equals the matching.

   König-এর উপপাদ্য বলে — bipartite graph-এ minimum vertex cover-এর আকার = maximum matching-এর আকার। নির্মাণ: matching বের করে unmatched বাঁ-পাশের nodes থেকে alternating BFS চালান, পৌঁছানো nodes Z হলে cover = (Left \ Z) ∪ (Right ∩ Z)।

3. Project selection: 4 projects with given profits and prerequisite tasks with given costs. Pick a subset to maximise profit minus cost.
   ৪টি project আছে — profit আছে, কিন্তু prerequisite task-এ খরচ আছে। সর্বাধিক net profit-এর জন্য কোন project বেছে নেবেন?
   ✨ Show Answer (উত্তর দেখুন)

   Reduction: Build a graph: source s → each project p with capacity = profit(p); each task t → sink t₀ with capacity = cost(t); each project → its required tasks with capacity ∞. Then answer = total_profit − min-cut. The min-cut splits projects into "do" (source side) and "skip" (sink side), and includes the cost of every task needed by selected projects.

   Reduction: source থেকে প্রতিটি project-এ profit capacity, প্রতিটি task থেকে sink-এ cost capacity, project থেকে required task-এ ∞ capacity। উত্তর = total_profit − min-cut।

4. Implement min-cut between s and t on the network from Figure 31.1 — print which nodes lie on the s-side.
   Figure 31.1-এর network-এ s ও t-এর মধ্যে min-cut বের করুন এবং s-পাশের nodes প্রিন্ট করুন।
   ✨ Show Answer (উত্তর দেখুন)

   ans4.cpp

   #include <bits/stdc++.h>
   using namespace std;
   const int N=6;
   int cap[N][N], flow[N][N];
   int bfs(int s,int t,vector<int>& p){fill(p.begin(),p.end(),-1);p[s]=s;queue<pair<int,int>> q;q.push({s,INT_MAX});while(!q.empty()){auto[u,f]=q.front();q.pop();for(int v=0;v<N;++v){int r=cap[u][v]-flow[u][v];if(p[v]==-1&&r>0){p[v]=u;int nf=min(f,r);if(v==t)return nf;q.push({v,nf});}}}return 0;}
   int main(){
       cap[0][1]=10;cap[0][2]=10;cap[1][3]=9;cap[1][4]=2;cap[2][4]=10;cap[3][5]=10;cap[4][5]=10;
       vector<int> p(N); int mf=0,push;
       while((push=bfs(0,5,p))>0){mf+=push;int v=5;while(v!=0){int u=p[v];flow[u][v]+=push;flow[v][u]-=push;v=u;}}
       // Now find S = reachable from 0 in residual
       vector<bool> vis(N,0);queue<int> q;q.push(0);vis[0]=1;
       while(!q.empty()){int u=q.front();q.pop();for(int v=0;v<N;++v)if(!vis[v]&&cap[u][v]-flow[u][v]>0){vis[v]=1;q.push(v);}}
       cout<<"Max flow = "<<mf<<"\nS-side nodes: ";
       for(int i=0;i<N;++i)if(vis[i])cout<<i<<" ";
       cout<<"\n";
   }

   Run Copy

5. Baseball elimination preview: with current standings and remaining games, decide if a team is eliminated.
   Cricket-এর Premier League-এ বর্তমান অবস্থান ও বাকি ম্যাচ থেকে কোনো দল title race থেকে eliminated কিনা — কীভাবে max-flow দিয়ে বের করবেন?
   ✨ Show Answer (উত্তর দেখুন)

   Idea: For target team x, give every other team y the maximum wins they can have without eliminating x: w[x] + r[x] − w[y]. Build a network: source s → each remaining game (i,j) with capacity g(i,j); each game → its two teams with capacity ∞; each team y → sink t with capacity (w[x]+r[x]−w[y]). If the max-flow saturates all source edges (= total remaining games among teams ≠ x), x can still win the title; otherwise x is eliminated. Used by ESPN-style standings as well as BPL projection systems.

   প্রতিটি অন্য দলকে x-কে না হারানোর সর্বোচ্চ জয় দিন। source → remaining game → দুই দল → sink — এই network-এ max-flow source-এর সব edge saturate করলে x এখনো জিততে পারে; নইলে eliminated।