Laser Phones
Given below code is for MLASERP spoj or laser phones spoj.
Explanation : -
First find the positions of the cow where they are situated . Now make one of them as source and other one destination . Now start bfs from source an apply bfs in following manner .
1-> for each cell having row = source.row or column = source.column make distance =0 as you need no mirror and push all that in queue.
2-> Next time whichever node you popped from queue , check if distance of all cell having row =poppedcell.row or column = poppedcell.column is greater then distance filled in poppedcell + 1 , if condition is true then push the node in queue .
Repeat above process till you reach the destination node.
#include <bits/stdc++.h> using namespace std; #define pb push_back #define mp make_pair #define pii pair<int , int > bool check[109][109]; int main(){ int n , m; cin>>m>>n; string s[n+9]; for(int i = 0 ; i < n ; i++) cin>>s[i]; pii source , dest; bool first = 1; for(int i = 0 ; i < n ; i ++){ for(int j = 0 ; j < m ; j++) if(s[i][j] == 'C'){ if(first) source = mp(i , j) , first = 0; else{ dest = mp(i , j); break; } } } queue<pair<int , int > > q; q.push(source); int res = 0; int dist[109][109]; for(int i = 0 ; i < 109 ; i++) for(int j = 0 ; j < 109 ; j++) dist[i][j] = INT_MAX; dist[source.first][source.second]=0; while(!q.empty()){ int i , j ; i = q.front().first; j = q.front().second; int u = i, v = j; while(u+1 < n && s[u+1][j]!='*' ){ if( dist[u+1][j] > dist[i][j] + 1){ q.push(mp(u+1 , j)); dist[u+1][j] = dist[i][j] +1; } u++; } u = i; while(u-1 >=0 && s[u-1][j]!='*' ){ if(dist[u-1][j] > dist[i][j] + 1){ q.push(mp(u-1 , v) ); dist[u-1][j] = dist[i][j] +1; } u--; } while(v+1 < m && s[i][v+1]!='*' ){ if( dist[i][v+1] > dist[i][j] + 1){ q.push(mp(i , v+1)); dist[i][v+1] = dist[i][j] +1; } v++; } v = j; while(v-1 >= 0 && s[i][v-1]!='*' ){ if(dist[i][v-1] > dist[i][j] + 1){ q.push(mp(i , v-1) ); dist[i][v-1] = dist[i][j] + 1; } v--; } q.pop(); } cout<<dist[dest.first][dest.second]-1<<endl; return 0; }
Please clarify the time complexity of the solution is it O(n^2).
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