10203 - Snow Clearing

[Even]

it says that...all roads are perfectly straight, with one lane in each direction ...

it means that every road has two lane ???
(from begin to end & from end to begin )

if it is, each graph's vertex should always have even degree, and Euler circuit always exist ...

if not, why sample output is 3:55?

I am confuse...can anybody help me? thx...

[Adrian Kuegel]

You should print the seconds with leading zeros (in C %02d).

[Even]

here I make two mistake ...

1. when min round to 60, it should be 00, and hour should add 1.

2. just like Adrian Kuegel says... min should has leading zero...
that is, ex) 3:5 ...you should show 3:05, not 3:5 or 03:05 ...

thank you ...Adrian Kuegel...

[razibcse]

I thot this is very easy...

just sum all the distances,double the distance and divide by 20...

but this is getting WA....

can anybody help me with this...

Code: Select all

#include <stdio.h>
#include <math.h>

void main()
{

float sum,x_start,x_end,y_start,y_end;
float x1,y1,x2,y2,dis,x_dif,y_dif,min_time,gap,dif;
long hour,minute,hanger_x,hanger_y,n,x,gap_dif;
scanf("%ld",&n);
for(x=0;x<n;x++)
  {

  scanf("%ld %ld",&hanger_x,&hanger_y);
  sum=0;
  while(scanf("%f %f %f %f",&x1,&y1,&x2,&y2)==4)
    {
    x_start=x1/1000;
    y_start=y1/1000;
    x_end=x2/1000;
    y_end=y2/1000;

    x_dif=x_end-x_start;
    if(x_dif<0)
      x_dif=(-1.0)*x_dif;

    y_dif=y_end-y_start;
    if(y_dif<0)
      y_dif=(-1.0)*y_dif;

    dis=sqrt((x_dif*x_dif)+(y_dif*y_dif));
    sum+=dis;
    }
   min_time=sum/10.0;

   hour=(long)min_time;

   dif=min_time-hour;

   gap=(min_time-hour)*60;
   gap_dif=(long)gap;

   if((gap-gap_dif)>0.00)
      minute=(long)(gap+1);
   else if((gap-gap_dif)==0.00)
      minute=(long)gap;
   printf("%ld:%ld\n",hour,minute);
 }
}

[/code]

[gvcormac]

I see at least two errors in your code.

In order to find them, consider all the inputs
that might yield an anwer of

3:00

[Cosmin.ro]

Even the official solution from Waterloo gets wa on this problem.

[Adrian Kuegel]

This problem is in multiple input format, so of course the waterloo solution gets WA because it expects other input format.
If you don't know multiple input format, read here: http://online-judge.uva.es/problemset/minput.html
Each problem with a blue checkmark or with a green checkmark is in multiple input format.

[zacharyleung]

I'm quite sure my code is correct, could anyone tell me what's wrong, cos I keep on getting WA....

[cpp]
#include <cstdio>
#include <cmath>
#include <cstring>

#define DEBUG(var) cout << #var << " = " << var << endl;
#define LINE cout << __LINE__ << endl;

const int MAX_LINE_LEN = 100;

void test();

int main() {
/*
test();
return 0;
*/

int n_cases;
double distance;
int x1, y1, x2, y2;
double time;
int hour, minute;
char buf[ MAX_LINE_LEN ];

scanf( "%d", &n_cases );
for( int case_num = 0; case_num < n_cases; case_num++ ) {
distance = 0;
scanf( "%d %d\n", &x1, &y1 );
while( fgets( buf, MAX_LINE_LEN, stdin ) != NULL ) {
if( buf[ 0 ] == '\n' )
break;
sscanf( buf, "%d %d %d %d", &x1, &y1, &x2, &y2 );
distance += sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
}

time = distance / 10000.0;
hour = (int) time;
minute = (int) ((time - hour) * 60 + 0.5);
if( minute == 60 ) {
minute = 0;
hour++;
}

if( case_num != 0 )
printf( "\n" );
printf( "%d:", hour );
if( minute < 10 )
printf( "0%d\n", minute );
else
printf( "%d\n", minute );
}

return 0;
}


/*
* Test values:
* 2.99 2:59
* 2.9999 3:00
* 3.001 3:00
* 3.01 3:01
*/
void test() {
float time;
int hour, minute;

time = 3.01;
hour = (int) time;
minute = (int) ((time - hour) * 60 + 0.5);

if( minute == 60 ) {
minute = 0;
hour++;
}
printf( "%d:", hour );
if( minute < 10 )
printf( "0%d\n", minute );
else
printf( "%d\n", minute );
}
[/cpp]

[Dominik Michniewski]

Think again - it's not good to use flating numbers ... exists simple solution which operates only on integers

Best regards
DM

[zacharyleung]

I tried to think about it, but I really have no idea how to do that. Since we use the square root of x^2 and y^2, won't that always be a float?

[Dominik Michniewski]

Sorry, I make wrong comment I see ...

My solution is mainly the same (counting distance) ... but I in different way compute time ...

Best regards
DM

[sidky]

I am a little confused about the problem. I thought this is like the 'Chinese postman problem'. But from the previous posts, i think i misunderstood. Can anyone please tell me, should we traverse each road twice, or at least once (like the postmans problem)?

[WR]

What's wrong with the following results?

input:

Code: Select all

3

0 0
0 0 10000 10000
5000 -10000 5000 10000
5000 10000 10000 10000

0 0
0 0 19999 0

10 10
-10000 -10000 10000 10000
10000 10000 -10000 10000
-10000 10000 -10000 -10000
-10000 -10000 10000 -10000
-298 -837 276 876
-10000 0 10000 0

output:

Code: Select all

3:55

2:00

11:01

And - if nothings's wrong, why do I get WA?

I've tried int, long and double as co-ordinate types (in C).

[Rony]

Hi,
Can any one help me,what's wrong with the code that's why i am getting
wrong answer ? Or, can any one give me some test cases?
Here is my code.


#include<stdio.h>
#include<stdlib.h>
#include<math.h>
double x,y,a,b,c,d,r,sum,v,sec,min,hour,integer;

int main(){

int tcase,n;
v=50.0/9.0;
//freopen("input.txt","r",stdin);

scanf("%d",&tcase);
while(tcase--){
scanf("%lf %lf\n",&x,&y);
sum=0.0;
while(scanf("%lf %lf %lf %lf\n",&a,&b,&c,&d)==4){
r=sqrt((a-c)*(a-c)+(b-d)*(b-d));
sum+=r*2;
}
min=(sum/v)/60.0;
hour=min/60.0;
min=modf(hour,&integer);
if((min*60.0)>0&&(min*60.0)<10) printf("%.0lf:0%.0lf\n",integer,(min*60.0)+.05);

else if((min*60.0)>59.0){

integer+=1;
min=0;
printf("%.0lf:0%.0lf\n",integer,min);
}

else printf("%.0lf:%.0lf\n",integer,(min*60.0)+.05);


putchar(10);

}
return 0;
}


Thanks.
"Still, I like her ."

[Guest]

Sorry my post was wrong !