UVa 12375 - Truchet Tiling

The tricky part of the solution is to realize how the color is flowing from cell to cell (regions).
We have two types of motifs 0 and 1. I have divided the motifs into 4 regions. In motif - 0 we have
regions 1,2,3,4 and in motif - 1 we have regions 5,6,7,8.
So,we will go from cell to cell and add the major region's area to solution and mark that region visited.
For region 1,4,6,7 we will add PI/4 (Pi*(r^2)/4) and for other regions we will add (4 - PI/2)/2 to
solution.
And about the movement , look at the figure we can't go to any cell from a certain cell
.For example,we can only go upward and left from region 1.Once We are in some cell if it's not
visited (if it's visited we have already added it's area to solution) and it's a valid cell(not out of the boundary) then we add that cell's/region's area to the
solution.

Code : -

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/**************************************** Cat Got Bored *****************************************/ #include <bits/stdc++.h> #define loop(i,s,e) for(int i = s;i<e;i++) //excluding end point #define pb(a) push_back(a) #define sqr(x) ((x)*(x)) #define CIN ios_base::sync_with_stdio(0); cin.tie(0); #define ll long long #define ull unsigned long long #define SZ(a) int(a.size()) #define read() freopen("input.txt", "r", stdin) #define write() freopen("output.txt", "w", stdout) #define ms(a,b) memset(a, b, sizeof(a)) #define all(v) v.begin(), v.end() #define PI acos(-1.0) #define pf printf #define sfi(a) scanf("%d",&a); #define sfii(a,b) scanf("%d %d",&a,&b); #define sfl(a) scanf("%lld",&a); #define sfll(a,b) scanf("%lld %lld",&a,&b); #define mp make_pair #define paii pair<int, int> #define padd pair<dd, dd> #define pall pair<ll, ll> #define fs first #define sc second #define CASE(t) printf("Case %d:\n",++t) // t initialized 0 #define INF 9999999999999 //10e9 #define EPS 1e-9 using namespace std; //Graph Construction int rowMove[9][3]; int colMove[9][3]; const int callRM[] = {0,0,1,1}; const int callCM[] = {0,1,0,1}; int RR,CC; int graph[205][205]; bool visited[205][205]; void graph_init() { rowMove[1][0] = 0; rowMove[1][1] = -1;rowMove[1][2] = 0; colMove[1][0] =-1; colMove[1][1] =0; colMove[1][2] =0; rowMove[2][0] = 0 ;rowMove[2][1] = -1 ;rowMove[2][2] = 1 ; colMove[2][0] = 1 ; colMove[2][1] = 0 ; colMove[2][2] = -1 ; rowMove[3][0] = 0 ;rowMove[3][1] = 1 ;rowMove[3][2] = -1 ; colMove[3][0] = -1;colMove[3][1] = 0;colMove[3][2] = 1; rowMove[4][0] = 0 ;rowMove[4][1] = 1 ;rowMove[4][2] = 0 ; colMove[4][0] =1;colMove[4][1] =0;colMove[4][2] =0; rowMove[5][0] = 0;rowMove[5][1] = -1;rowMove[5][2] = 1; colMove[5][0] = -1;colMove[5][1] = 0;colMove[5][2] = 1; rowMove[6][0] = 0;rowMove[6][1] = -1;rowMove[6][2] = 0; colMove[6][0] =1;colMove[6][1] =0;colMove[6][2] =0; rowMove[7][0] =0;rowMove[7][1] =1;rowMove[7][2] =0; colMove[7][0] = -1;colMove[7][1] = 0;colMove[7][2] = 0; rowMove[8][0] = 0;rowMove[8][1] = 1;rowMove[8][2] = -1; colMove[8][0] = 1;colMove[8][1] = 0;colMove[8][2] = -1; } bool valid(int r,int c) { if(r<1 || r>2*RR || c<1 || c>2*CC) return false; return true; } double area_grid(int r,int c) { int grid_no = graph[r][c]; int g_n = grid_no; if(g_n == 1 || g_n == 4 || g_n == 6 || g_n == 7 ) return ((PI)/4.00000)*1.00000; else return (4.0000000 - (PI/2.000000)*1.000000)/2.0000000; } double area(int r,int c) { if(visited[r][c]) return 0.00000000; if(!valid(r,c)) return 0.00000000; visited[r][c] = true; double partial_area = 0.00000000; for(int i=0; i<3; i++) { int r_pp = r + rowMove[ graph[r][c] ][i]; int c_pp = c + colMove[ graph[r][c] ][i]; partial_area += area(r_pp,c_pp); } return area_grid(r,c) + partial_area ; } int main() { int tc; cin>>tc; int cas = 0; graph_init(); while(tc--) { cin>>RR>>CC; int rrr = 1 ; for(int r = 0; r<RR; r++) { string sss; cin>>sss; int ccc = 1 ; for(int c = 0; c<CC; c++) { if(sss[c]=='0') { int r_p = 2*r + 1; int c_p = 2*c + 1; graph[r_p][c_p] = 1; graph[r_p][c_p+1] = 2; graph[r_p+1][c_p] = 3; graph[r_p+1][c_p+1] = 4; } if(sss[c]=='1') { int r_p = 2*r + 1; int c_p = 2*c + 1; graph[r_p][c_p] = 5; graph[r_p][c_p+1] = 6; graph[r_p+1][c_p] = 7; graph[r_p+1][c_p+1] = 8; } } } /* int i,j; loop(i,1,2*RR+1) { loop(j,1,2*CC+1) { cout<<graph[i][j]; } cout<<endl; } */ //calling inititialization CASE(cas); int Q; cin>>Q; while(Q--) { ms(visited,false); int r_call , c_call; cin>>r_call>>c_call; double ans = 0.00000000; if((r_call%2==0 && c_call%2==1) || (r_call%2==1 && c_call%2==0)) { pf("%.4f\n",ans); continue; } int flag = 0; for(int ii= 0; ii<4; ii++) { int r_t = r_call + callRM[ii]; int c_t = c_call + callCM[ii]; if(!valid(r_t,c_t)) continue; if(graph[r_t][c_t]==2 || graph[r_t][c_t]==3 ||graph[r_t][c_t]==5 ||graph[r_t][c_t]==8) { ans = area(r_t,c_t); flag = 1; break; } } if(flag == 1) pf("%.4f\n",ans); else { for(int ii= 0; ii<4; ii++) { int r_t = r_call + callRM[ii]; int c_t = c_call + callCM[ii]; if(valid(r_t,c_t)) { ans = area(r_t,c_t); pf("%.4f\n",ans); break; } } } } } return 0; }

All palindromic contiguous substring !

This is so far what I have come up with ! I think there are better ways to do this .  : ) )
But it 's working for my problem
:-P


Here's to the code - >

//Sorry this doesn't work ! I will edit it when I'm busy :-/

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/* @_@ catGotBored @_@ */ #include<bits/stdc++.h> using namespace std; int isPalindrome[100][100]; map<string,int>isThere; void all_palindromic_conti_sub_str(string s) { int N = s.length(); //Lets find all such palindromes with length 1 . Easy , huh ? for(int idx = 0; idx<N; idx++) isPalindrome[idx][idx] = 1; //Lets find all such palindromes with length 2 . Easy again !! for(int idx = 0; idx < N -1 ; idx++) isPalindrome[idx][idx+1] = (s[idx] == s[idx+1] ); //Now,lets do it for all lengths < = N //Idea is simple : //If I know s[i...j] is a palindrome then s[i-1.....j+1] is a palindrome //if and only if s[i-1] == s[j+1] //O(N^2) for marking only for(int idx = 0 ,len = 3; idx <= N -len; idx++,len++) { for(int jdx = idx + len -1 ; jdx < N ; jdx ++ ) { isPalindrome[idx][jdx] = isPalindrome[idx+1][jdx-1] & s[idx] == s[jdx] ; } } / //Bad approach : O(N^3 * log (N)) -> Need to do better // for(int ff = 0; ff<N; ff++) for(int ss = 0; ss<N; ss++) if(isPalindrome[ff][ss]) { string temp; for(int xfsx = ff ; xfsx <= ss ; xfsx++) temp+=s[xfsx] ; // cout<<s[xfsx]; //Every palindromes non distinct ones too if(isThere[temp] == 0) { cout<<temp; isThere[temp]++; cout<<endl; } } } int main() { string str_ = "aassaa"; memset(isPalindrome,0,sizeof(isPalindrome)); // just for nothing all_palindromic_conti_sub_str(str_); return 0; }