Moderator: Board moderators
interesting. i'll think about it.
Code: Select all
#include <iostream.h>
#include <string.h>
#include <stdio.h>
void main() {
int i,j,g,p,x,y;
char niz[35][35],a[200],last;
cin >> g;
for (p=0;p<g;p++) {
cin >> x >> y;
y = 31 - y; // da bi prebacili y na donji deo grafika...
for (i=0;i<32;i++) // init grafika
for (j=0;j<32;j++) niz[i][j]='.';
//gets(a);
// my NEW input...
cin.getline( a, 200);
if(strlen(a)==0) cin.getline( a, 200);
last = 'z';
// ############# :) ##############
//if (a[0]=='S') y++; // HAAAAR HAR HAR ... :)
for (j=0;j<(signed)strlen(a)-1;j++) {
if (a[j]=='E') {
if (last=='E') x++;
else if (last=='N') { x++; y--; }
niz[y+1][x]='X';
last = 'E';
//if (a[j+1]=='E') x++;
//if (a[j+1]=='S') {y++; x++;}
}
if (a[j]=='N') {
if (last=='N') y--;
else if (last=='W') { x--; y--; }
niz[y][x+1]='X';
last = 'N';
//niz[y][x+1]='X';
//if (a[j+1]=='N') y--;
//if (a[j+1]=='E') {y--; x++;}
}
if (a[j]=='W') {
if (last=='W') x--;
else if (last=='S') { x--; y++; }
niz[y-1][x]='X';
last = 'W';
//niz[y-1][x]='X';
//if (a[j+1]=='W') x--;
//if (a[j+1]=='N') {y--; x--;}
}
if (a[j]=='S') {
if (last=='S') y++;
else if (last=='E') { x++; y++; }
niz[y][x-1]='X';
last = 'S';
//niz[y][x-1]='X';
//if (a[j+1]=='S') y++;
//if (a[j+1]=='W') {y++; x--;}
}
} // end for j
cout << "Bitmap #" << p+1 << endl;
// output
for (i=0;i<32;i++) {
for (j=0;j<32;j++) cout << niz[i][j];
cout << endl;
}
cout << endl;
} // kraj glavnog for p
}
Code: Select all
if (a[j]=='E') { niz[y+1][x++]='X'; }
if (a[j]=='N') { niz[y--][x]='X'; }
if (a[j]=='W') { niz[y][--x]='X'; }
if (a[j]=='S') { niz[++y][x-1]='X'; }
seriously. I referred to the mathematical coordinate system, which is shown in the problem's statement. By the box I meant the square with corners at the points (x,y), (x,y-1), (x-1,y-1), (x-1,y). When the point moves from (x,y) down to (x,y-1), this square is always part of the border. And thus, the program should put 'X' in the array's cell, which corresponds to it. It depends on the programmer, what exactly the array represents, after all.one more thing, MF, you said:
"....and the border is always a "box" whose right side is the segment ((x,y) - (x,y-1)).... "
i don't get it - what exactly is the 'segment' ? by "box" i guess you mean one character place in my 2D array (char[32][32]), but what is "segment ((x,y) - (x,y-1)) " ?
haha..,Code: Select all
scanf ("%d",&g); // for reading number of test cases
scanf ("%d%d\n",&x,&y); // for reading starting coordinates
gets(a); // for reading the whole line of commands.
Code: Select all
//
// File: newmain.cc
// Author: Vaio
//
// Created on September 17, 2008, 7:32 PM
// There is a difference between points and squares but the diagram is plain
// wrong.
#include <iostream>
#include <string>
#include <sstream>
#include <limits>
using namespace std;
enum { north, east, south, west };
struct Point
{
int x, y;
Point() : x(0), y(0) { }
Point(int xx, int yy) : x(xx), y(yy) { }
Point(const Point& p) : x(p.x), y(p.y) { }
friend ostream& operator<<(ostream& o, const Point& p)
{
return o << "[" << p.x << "," << p.y << "]";
}
Point& go(const int i)
{
switch(i)
{
case west: x--; break;
case east: x++; break;
case north: y++; break;
case south: y--; break;
}
return *this;
}
};
struct Square
{
bool n, e, s, w;
Square() { n = e = s = w = false; }
bool isEmpty() const
{
if(!n && !e && !s && !w) return true;
return false;
}
friend ostream& operator<<(ostream& o, const Square& s)
{
return o << boolalpha << s.n << " : " << s.e << " : " << s.s << " : " << s.w;
}
};
class Bitmap
{
private:
Square sq[32][32];
Point begin;
string inputs;
public:
Bitmap(const Point& p) : begin(p) { }
void draw()
{
Point cur(begin);
for(string::iterator it = inputs.begin(); it != inputs.end(); it++)
{
char ch = *it;
Point rhs(cur);
Point origin;
sq[origin.y][origin.x].n = true; // the origin.
if(ch == '.') break;
switch(ch)
{
case 'N':
rhs.x++;
sq[rhs.y][rhs.x].w = true;
cur = cur.go(north);
break;
case 'E':
rhs.y--;
sq[rhs.y][rhs.x].n = true;
cur = cur.go(east);
break;
case 'W':
rhs.y++;
sq[rhs.y][rhs.x].s = true;
cur = cur.go(west);
break;
case 'S':
rhs.x--;
sq[rhs.y][rhs.x].e = true;
cur = cur.go(south);
break;
}
#ifdef DEBUG
cout << "cur: " << cur << '\t' << "rhs: " << rhs << endl;
#endif
}
}
friend ostream& operator<<(ostream& o, const Bitmap& b)
{
for(int y = 31; y >= 0; y--)
{
o << endl;
for(int x = 0; x < 32; x++)
{
if(b.sq[y][x].isEmpty()) o << '.';
else if(y == 0 && x == 0) o << 'O'; // bearing check
else o << 'X';
}
}
return o;
}
friend istream& operator>>(istream& i, Bitmap& b)
{
return i >> b.inputs;
}
};
int main(int argc, char** argv) {
int n;
cin >> n; // Check for numeric limits!
numeric_limits<int> lim;
if(n >= lim.max())
{
cerr << endl << "The monkey wrench! Argh!!!" << endl;
return 1;
}
for(int i = 0; i < n; i++)
{
cout << endl << "Bitmap #" << i+1 << endl;
int startx, starty;
cin >> startx >> starty;
Point p(startx, starty);
Bitmap bt(p);
cin >> bt;
bt.draw();
cout << bt << endl;
}
return (EXIT_SUCCESS);
}