I didn't look into the correctness of the algorithm.
But I can tell you the reason for that "compilation error"
Visual Studio 6.0 does not follow ANSI standards!
Especially the scope of the loop variable 'i' is the thing which causes compilation error for you.
In ANSI C (or gcc for that matter),
if you declare 'i' like this,
for(int i=0;i<n;i++)
{
}
}
outside the loop you cannot access it. Its scope is limited within the loop.So if you want to declare another loop, again you have to say,
for(int i=0;in;i++)
{
}
Suggestions:
1. If possible use some gcc based compilers available for windows (There are mny available for windows).
2. Try cygwin
3. Use Visual Studio 2003 or 2005 (The 'bug' is fixed in these version ;))
Regards,
Prunthaban
On 8/26/06, Iman <[EMAIL PROTECTED]> wrote:
today i was solving this problem
http://acm.uva.es/p/v101/10189.html
and i after i solve it i compile it with Microsoft Visual Studio 6 and
i've got the results
correct but
after i send it to Valladolid judge it say it has a compiler error!!! o
my god
anyway need your helps
tanx in advance
------
/*
problem No. 10189 of the Valladolid
the problem gets input as a mine field
it gets mine fields in this way
first it gets two numbers that first integer is number of lines
the second integer is number of columns and the program stops when
the user inputs these both two numbers 0
after getting this N and M it gets a N*M matrix of characters
in this matrix the character '*' defines a mine and character '.'
define empty space
Output :
Field #Number of field that is going to being processed
next lines is a M*N matrix that '*' characters will apear in the
output
exactly like inputs and instead of empty space we put numbers in those
cells the numbers define numbers of mines cross that cell
obviously the minimum is 0 and the Maximum could be is 8
*/
#include <iostream.h>
void checker(int f1[][100],int m,int n)
{
int Currtemp = 0; // Number of mines across the current cell should
being zero at first of counting for each cell
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
Currtemp = 0;
if(f1[i][j]!=-1)
{
if(f1[i-1][j-1] == -1) //1
Currtemp++;
if(f1[i-1][j] == -1) //2
Currtemp++;
if(f1[i-1][j+1] == -1) //3
Currtemp++;
if(f1[i][j-1] == -1) //4
Currtemp++;
if(f1[i][j+1] == -1) //5
Currtemp++;
if(f1[i+1][j-1] == -1) //6
Currtemp++;
if(f1[i+1][j] == -1) //7
Currtemp++;
if(f1[i+1][j+1] == -1) //8
Currtemp++;
f1[i][j] = Currtemp;
}
}
}
}
int main()
{
int RawField[100][100] = {0}; // Our raw mine field
int field[100][100] = {0}; // Our mine field
int NOF = 0; // Number of Current field
int m = 0; // Number of lines
int n = 0; // Number of columns
char temp; //this help us to convert chars to int
cin>>m>>n;
while((m!=0)||(n!=0))
{
NOF++;
for(int i=0;i<100;i++)
{
for(int j=0;j<100;j++)
RawField[i][j] = 0;
}
for(i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
cin>>temp;
if(temp=='*')
{
RawField[i][j] = -1;
}
}
}
cout<<"Field #"<<NOF<<":"<<endl;
checker(RawField,m,n);
for(i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(RawField[i][j] !=-1)
cout<<RawField[i][j];
else
cout<<'*';
}
cout<<endl;
}
cout<<endl;
cin>>m>>n;
}
return 0;
}
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