Theory:
This is an np-hard problem. As the degree of each node can be at most 3 so using dfs ,it can be solved.
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#include <iostream> #include <stdio.h> #include <vector> #include <string.h> using namespace std; //vector<int> vb[30]; int mat[30][30]; int visit[30]; int n,m; int length; void search(int a,int len) { if(len>length) length = len; for(int i=0;i<n;i++) { //a=vb[step][i]; if(mat[a][i]>0) { mat[a][i]-=1; mat[i][a]-=1; search(i,len+1); mat[a][i]++; mat[i][a]++; } } } int main() { int a,b; //memset(visit,0,sizeof(visit)); while(cin>>n>>m) { if(n==0 && m==0) break; memset(mat,0,sizeof(mat)); //for(int i=0;i<n;i++) //vb[i].clear(); length=0; for(int i=0;i<m;i++) { cin>>a>>b; //vb[a].push_back(b); //vb[b].push_back(a); mat[a][b]++; mat[b][a]++; } int len =0; for(int i=0;i<n;i++) { //visit[0]=1; search(i,0); //visit[0]=0; } cout<<length<<"\n"; } return 0; } |