C++ || Using BFS
As its mentioned in first Solved example that for any two Bombs A-B , B will blast due to effect of A if and only if Centre of B lies inside or on the circumfer
As its mentioned in first Solved example that for any two Bombs A-B , B will blast due to effect of A if and only if Centre of B lies inside or on the circumference of A  => radius of Bomb A (r1) >= Distance between their centers.\n\nSo We start bfs from Bomb i and including all those bombs in effect of A that satisfies above condition.\n\n```\n#define lln long long int\nclass Solution {\n  int mx=INT_MIN;\n  queue<int>q;\n\npublic:\n    int maximumDetonation(vector<vector<int>>& bombs) {\n        \n      int n=bombs.size();\n      \n      for(int i=0;i<n;i++){\n         bfs(i,bombs);\n      }\n      \n      return mx;\n    }\n  \n      void bfs(int i, vector<vector<int>>& bombs){\n          int n=bombs.size();\n        vector<bool>vis(n,false);\n        \n         lln x1,y1,r1,x2,y2,r2;\n          q.push(i);\n          vis[i]=true;\n        \n         int count=1;\n        \n        while(!q.empty()){\n    \n          int j=q.front();\n          vis[j]=true;\n          q.pop();\n          \n          for(int k=0;k<n;k++){\n            if(vis[k]==true)\n              continue;\n            \n      x1=bombs[j][0],y1=bombs[j][1],r1=bombs[j][2];\n      x2=bombs[k][0],y2=bombs[k][1],r2=bombs[k][2];\n          \n            if(r1 >= sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)) ) {\n              vis[k]=true;\n               q.push(k);\n              count++;\n          }\n        }\n          \n      }\n               \n       mx=max(mx,count);\n    }\n};\n```
['Breadth-First Search']