#Region "Change Background Color of MDIParent : bgcolor()" 'to change the background color of mdiparent 'for more : http://acomputerengineer.wordpress.com Private Sub bgColor() Dim child As Control For Each child In Me.Controls If TypeOf child Is MdiClient Then child.BackColor = Color.CadetBlue Exit For End If Next child = Nothing End Sub #End Region
Kill the code
Sunday, 22 September 2013
Change background color of MDIParent Form in VB.Net
Sunday, 31 March 2013
Rope Intranet : Google Code Jam
Problem StatementProblem A company is located in two very tall buildings. The company intranet connecting the buildings consists of many wires, each connecting a window on the first building to a window on the second building. You are looking at those buildings from the side, so that one of the buildings is to the left and one is to the right. The windows on the left building are seen as points on its right wall, and the windows on the right building are seen as points on its left wall. Wires are straight segments connecting a window on the left building to a window on the right building. You've noticed that no two wires share an endpoint (in other words, there's at most one wire going out of each window). However, from your viewpoint, some of the wires intersect midway. You've also noticed that exactly two wires meet at each intersection point. On the above picture, the intersection points are the black circles, while the windows are the white circles. How many intersection points do you see? Input The first line of the input gives the number of test cases, T. T test cases follow. Each case begins with a line containing an integer N, denoting the number of wires you see. The next N lines each describe one wire with two integers Ai and Bi. These describe the windows that this wire connects: Ai is the height of the window on the left building, and Bi is the height of the window on the right building.
Rope Intranet : Google Code Jam 
Friday, 29 March 2013
Water Shield : Google Code Jam
PROBLEM DESCRIPTION
:
Geologists sometimes
divide an area of land into different regions based on where rainfall flows
down to. These regions are called drainage basins.
Given an elevation map
(a 2dimensional array of altitudes), label the map such that locations in the
same drainage basin have the same label, subject to the following rules.
 From each cell, water flows
down to at most one of its 4 neighboring cells.
 For each cell, if none of its 4
neighboring cells has a lower altitude than the current cell's, then the
water does not flow, and the current cell is called a sink.
 Otherwise, water flows from the
current cell to the neighbor with the lowest altitude.
 In case of a tie, water will
choose the first direction with the lowest altitude from this list: North,
West, East, South.
Every
cell that drains directly or indirectly to the same sink is part of the same
drainage basin. Each basin is labeled by a unique lowercase letter, in such a
way that, when the rows of the map are concatenated from top to bottom, the
resulting string is lexicographically smallest. (In particular, the basin of
the most NorthWestern cell is always labeled 'a'.)
SOLUTION:
public class water_shield1 { static int val1 = 0,row = 3,col = 3; /*static int input[][] = {{1,2,3,4,5},{2,9,3,9,6},{3,3,0,8,7},{4,9,8,9,8},{5,6,7,8,9}}; /*static int input1[][] = {{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0}}; static int input2[][] = {{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0}};*/ static int input[][] = {{9,6,3},{5,9,6},{3,5,9}}; /*static int input[][] = {{1,2,3},{4,5,6},{7,8,9}};*/ static int input1[][] = {{0,0,0},{0,0,0},{0,0,0}}; static int input2[][] = {{0,0,0},{0,0,0},{0,0,0}}; /*static int input[][] = {{8,8,8,8,8,8,8,8,8,8,8,8,8},{8,8,8,8,8,8,8,8,8,8,8,8,8}}; static int input1[][] = {{0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0}}; static int input2[][] = {{0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0}};*/ public static void main(String args[]) { //water_shield1 w = new water_shield1(); int max = input[0][0],min = input[0][0]; int i1 = 0,j1 = 0,count = 0; for(int i=0;iinput[i][j]) { min= input[i][j]; } } } for(int t=0;t
=0) { min_ard = input[i][j1]; a = i; b = j1; } if((i1)>=0 && min_ard > input[i1][j]) { min_ard = input[i1][j]; a = i1; b = j; } if((i+1)
input[i+1][j]) { min_ard = input[i+1][j]; a = i+1; b = j; } if((j+1)
input[i][j+1]) { min_ard = input[i][j+1]; a = i; b = j+1; } //System.out.println("a = " + a + " b = " + b); //System.out.println("input1[a][b]" + input1[a][b] + " input1[i][j] = " + input1[i][j]); if(input1[a][b] == 0) { if(input1[i][j] != 0) { input1[a][b] = input1[i][j]; //System.out.println("assign..."); //replace(input1[a][b], input1[i][j]); } else { input1[a][b] = input1[i][j] = ++val1; //System.out.println("val1" + val1 + " i=" + i + " j = " + j + " a = " + a + " b = " + b); } } else if(input[a][b] != input[i][j]) { //System.out.println("input[i][j] = " + input1[i][j] + "input[a][b] = " + input1[a][b]); int temp = input1[i][j]; input1[i][j] = input1[a][b]; if(input1[a][b] < temp) replace(temp, input1[a][b]); else replace(input1[a][b],temp); //System.out.println("replace"); } else { input1[i][j] = ++val1; //System.out.println(""); } } static void replace(int a,int b) { //a is original value, to be replaced with b. for(int i=0;i

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