UVA 10110 - Light, more light

In this problem there will be a number given.

we had to find if the n'th light is on or off? if a light is toggle odd number of time than it is on otherwise its off.

Observations: a number has odd number of factors if and only if it is a perfect square.

Code:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

#include<bits/stdc++.h> using namespace std; #define ll long long int main(){ ll n; while(1){ cin>>n; if(n==0){ break; } ///just has to check n is a perfect square or not? because all of the perfect square only ///has odd number of factors. and if number of factors is odd then it will be on. ll xx= sqrt(n); if(xx*xx==n){ cout << "yes\n"; } else cout << "no\n"; } return 0; }

A simple Cycle detection Problem using DFS :: Codeforces 977E

 Problem Link

In this problem several connected components are given, we need to find number of such components with cycle.

A components is contain a cycle if and only if all of its vertices has degree exactly 2.

To find connected components of the graph, we will use  a simple dfs through unvisited nodes. You can learn dfs from here

After that we will check if all the vertices has degree exactly 2 or not.

Official Editorial

Code:

 

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

#include<bits/stdc++.h> using namespace std; #define pb push_back ///the connected component is a cycle iff the degree of each its vertex equals to 2 const int maxn=200005; vector<int>edge[maxn],connected; int vis[maxn],degree[maxn]; int cnt=0; void dfs(int u){ vis[u]=1; connected.pb(u); for(auto it:edge[u]){ if(vis[it]==0){ dfs(it); } } } int main(){ int n,m; cin>>n>>m; for(int i=0;i<m;i++){ int u,v; cin>>u>>v; edge[u].pb(v); edge[v].pb(u); degree[u]++; degree[v]++; } for(int i=1;i<=n;i++){ if(vis[i]==0){ connected.clear(); dfs(i); bool ok=1; for(auto it:connected){ if(degree[it]!=2){ ok=0; } } if(ok){ cnt++; } } } cout << cnt << endl; }

Guilty Prince Lightoj-1012

Problem link

This problem is straightforward implementation of floodfill algorithm.

Code:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67

#include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mp make_pair #define IOS ios::sync_with_stdio(false); int dx4[] = {0, 0, 1, -1}; ///direction arrays int dy4[] = {1, -1, 0, 0}; const int maxn=25; char s[maxn][maxn]; map<pair<int,int>,int>vis; int w,h; int ans; void init(){ ans=0; vis.clear(); } void bfs(int posi,int posj){ pair<int,int>x={posi,posj}; queue<pair<int,int> > q; q.push(x); vis[x]=1; while(!q.empty()){ pair<int,int>u=q.front(); q.pop(); int x=u.first; int y=u.second; for(int i=0;i<4;i++){ int nx=x+dx4[i]; int ny=y+dy4[i]; if(nx>=0 && nx<h && ny>=0 && ny<w && s[nx][ny]=='.' && vis[mp(nx,ny)]==0){ ans++; vis[mp(nx,ny)]++; q.push(mp(nx,ny)); } } } } int tc=0; ///solution void solution(){ cin>>w>>h; int posi,posj; init(); for(int i=0;i<h;i++){ for(int j=0;j<w;j++){ cin>>s[i][j]; if(s[i][j]=='@'){ posi=i; posj=j; } } } bfs(posi,posj); cout<< "Case "<< ++tc <<": "<< ans+1 << endl; } signed main() { IOS int t; t=1; cin>>t; while(t--){ solution(); } return 0; }