This is a bipartite matching problem. This can be solved using Ford fulkerson algorithm for Max flow problem.
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#include <iostream> #include <limits.h> #include <string.h> #include <queue> using namespace std; int gr[500][500]; int rgr[500][500]; string str[50]; int h,w; int numOfantena; int parent[500]; bool visited[500]; int src,targt; bool bfs(int s, int t) { memset(visited, 0, sizeof(visited)); queue <int> q; q.push(s); visited[s] = true; parent[s] = -1; while (!q.empty()) { int a = q.front(); q.pop(); for (int i=0; i<=targt; i++) { if (visited[i]==false && rgr[a][i] > 0) { q.push(i); parent[i] = a; visited[i] = true; } } } return (visited[t] == true); } int fordFulkerson(int s, int t) { int u, v; for (u = 0; u <=targt; u++) for (v = 0; v <= targt; v++) rgr[u][v] = gr[u][v]; int max_flow = 0; memset(parent, -1, sizeof(parent)); while (bfs(s, t)) { int path_flow = INT_MAX; for (v=t; v!=s; v=parent[v]) { u = parent[v]; path_flow = min(path_flow, rgr[u][v]); } for (v=t; v != s; v=parent[v]) { u = parent[v]; rgr[u][v] -= path_flow; rgr[v][u] += path_flow; } max_flow += path_flow; } return max_flow; } int main() { int t; cin>>t; while(t--) { cin>>h>>w; numOfantena = 0; for(int i=0;i<h;i++) { cin>>str[i]; } memset(gr,0,sizeof(gr)); memset(rgr,0,sizeof(rgr)); src = 0; targt = h*w+1; for(int i=0;i<h;i++) { for(int j=0;j<w;j++) { if(str[i][j]=='*') { numOfantena++; if((i+j)%2) { gr[src][i*w+j+1]=1; if(i-1>=0 && str[i-1][j]=='*') { gr[i*w+j+1][(i-1)*w+j+1]=1; } if(i+1<h && str[i+1][j]=='*') { gr[i*w+j+1][(i+1)*w+j+1]=1; } if(j-1>=0 && str[i][j-1]=='*') { gr[i*w+j+1][i*w+j-1+1]=1; } if(j+1<w && str[i][j+1]=='*') { gr[i*w+j+1][i*w+j+1+1]=1; } } else gr[i*w+j+1][targt]=1; } } } int flow = fordFulkerson(src, targt); cout<<(numOfantena-2*flow)+flow<<"\n"; } return 0; } |