Depth First Search DFS (code (C++)

This algorithm can transverse a graph and find all the reachable points

 1  #include<bits/stdc++.h>

 2  using namespace std;

 3  int n,e;

 4  vector<int>g[128];

 5  bool seen[128];

 6 

 7  void dfs(int u){

 8  seen[u] = true;

 9  cout<<u<<" ";

10  for(int i=0;i<g[u].size();i++){

11      int v = g[u][i];

12      if(!seen[v])dfs(v);

13  printf("\n");

14  }

15 

16  }

17  int main(){

18  cin>>n>>e;

19  for(int i=0;i<n;i++){

20      int u,v;

21      cin>>u>>v;

22      g[u].push_back(v);

23  }

24  for(int i = 1;i<=n;i++ ){

25      if(!seen[i]){

26          dfs(i);

27      }

28  }

29 

30  return 0;}

UVA Problem 10685 - Nature (C++)

The question to the problem can be found here:

Explanation:
1. You need to declare two Array, one for Parent and other for number of Children in this case as chains of animals.
2. You need to use MAP data structure to store names as integers.
3. First of all Parents of Node are itself.
4. Take the Names(C) as input Node and Names.
5. Take Relationships between two elements as edges. Now the logical part.
6. If parents of these edges don't match set the 1st name(corresponding int) parent of the 2nd name(corresponding int).
7. Add the array of 2nd to the array of 1st to keep track of size of chain
8. Find the maximum element in the array
9.Happy Coding :)

Code:
#include<bits/stdc++.h>
using namespace std;
int par[5002],arr[5002];
map<string,int>mp;
int find(int n){
if(par[n] == n)
    return n;
return find(par[n]);
}
void check(int u,int v){
int U,V;
U=find(u);
V=find(v);
//cout<<U<<" "<<V<<endl;
if(U!=V){
    par[V]=U;
    arr[U]+=arr[V];
}
}
void makeset(int n){
for(int i=0;i<=n;i++)
par[i]=i;
}

int main(){

string s1,s2,s3;
int node,edge;
while(scanf("%d%d",&node,&edge)==2){
    if(node == 0 && edge ==0){
        break;
    }

    makeset(node);
    for(int i=0;i<node;i++){
        arr[i] = 1;
    }
    for(int i=0;i<node;i++){

        cin>>s1;
        mp[s1]=i;
    }

    for(int i=0;i<edge;i++){
        cin>>s2>>s3;
        int x= mp[s2];
        //cout<<x<<endl;
        int y= mp[s3];
       // cout<<y<<endl;
        check(x,y);
    }
int maximum =0;
for(int i=0;i<node;i++){
    maximum = max(maximum,arr[i]);
}
cout<<maximum<<endl;
}

return 0;}

Uva Problem 544 - Heavy Cargo Solution (C++)

Welcome:

The question to the problem can be found here:


Explanation:

1. The truck size is not needed since the problem can be solved using the lowest cost of path.
2. Use Map function to dynamically point city name(string) to Index(int).
3.++Index means value is increased and then assigned.
4. Use a Modified Floyd Warshal Algorithm as shown below.
5. Every Node in the array d now point to the minimum cost.
6. Finally the source to destination of the array is printed.
7.Happy Coding

Code:

#include<bits/stdc++.h>
using namespace std;
#define clr(a,replace) memset(a,replace,sizeof(a))
#define INF 99999
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define PB push_back
#define MAX 100000
#define LL long long
#define pf printf

int Case=0;
int main(){
//freopen("input.txt","r",stdin);
//   freopen("output.txt","w",stdout);
int n,r;

while(cin>>n>>r){
        if(n==0 && r ==0){
            return 0;
        }

    map<string,int> city;

    string x,y;
    int cost;
    int index=0;
    int d[250][250];
    rep(i,250){
    rep(j,250){
    d[i][j]=-1;
if(i==j){
    d[i][j] =0;
}
    }}
    for(int i=0;i<r;i++){
            cin>>x>>y>>cost;
        if(!city[x]){
            city[x] = ++index;
        }
        if(!city[y]){
            city[y]= ++index;
        }
    d[city[x]][city[y]] =cost;
    d[city[y]][city[x]] =cost;
    }

    for(int k=1;k<=n;k++){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
           d[i][j]=d[j][i] = max(d[i][j],min(d[i][k],d[k][j]));
        }
    }
    }
Case++;
    string source;
    string destination;
    cin>>source>>destination;
    pf("Scenario #%d\n%d tons\n\n",Case,d[city[source]][city[destination]]);

    city.clear();

}


return 0;}