10245 - The Closest Pair Problem

[yatsen]

I got Time Limit Exceeded for this problem.
I use 2 for loop to caculate all pair distance.
Can anyone tell me what's wrong?

[junjieliang]

Look at the limit: N <= 10000. Given a O(n^2) method you'll never get it to run in time.

Btw, the quote at the bottom fits exactly into this situation...

[yatsen]

[Stefan Pochmann]

Go for O(n log n). Get yourself the book of Cormen, Leiserson, Rivest (and Stein in the 2nd edition), if you want the algorithm explained. It's not the easiest algorithm I've ever seen...

[Stefan Pochmann]

Btw, in a local contest we once had a similar problem and I wrote a fake because we didn't have enough time. I splitted the world into a grid and only checked pairs in the same or in adjacent cells. Didn't work here, unfortunately. But back then, it did. There are two extremes. If you make the cells too large, you'll end up with n^2 and runtime-limit exceeded again. If you make them too small, you'll get wrong answer. But somewhere in between may lie an "accepted"

[Ivan Golubev]

Actually O(n^2) isn't so large as you think. So this problem can be solved only with stupid brute-force, as I did. It rans for 19 seconds but it's enough to fit into time limit. Of course, while checking distance don't use sqrt. It must be used only once, before outputting result.

[Stefan Pochmann]

Yeah, right. But can you live with the shame of having such a naive solution?

[shahriar_manzoor]

Actually it's time limit was 8 seconds in the contest, and its judge data was prepared accordingly. But I am afraid that the OJ decides to keep time limit for all problems as 30 seconds in 24-hour judge. So I will have to come up with new idea! more judge datas

<font size=-1>[ This Message was edited by: shahriar_manzoor on 2002-03-21 12:14 ]</font>

[Caesum]

After much messing with this problem I seem to be timing out on trying to parse the input file. Am I right in saying that the input is full of non-numeric characters as my usual scanf("%u",&n) just seems to hang. In the end I have written something based on getc(stdin) hunting for numeric digits, but now some of the arrays seem to be bigger then 10000 as stated in the question. Is there something funny going on here or am I missing something ?

[Ivan Golubev]

Hmm, there nothing about 'only integers in input' in the problem statement. Why not to use scanf("%lf", &x) ?

[M.Z]

my method is O(n*log(n)) But Wrong answer.
any one can help me!
thank you!

[cpp]
#include<iostream.h>
#include<stdio.h>
#include<math.h>

#include<fstream.h>
ifstream in("es_10245.in");
//#define in cin

const double eps=1e-14;
struct point{
double x,y;
} p[10000];
int N;


void swap(int i,int j)
{
double temp;
temp=p[i].x;
p[i].x=p[j].x;
p[j].x=temp;

temp=p[i].y;
p[i].y=p[j].y;
p[j].y=temp;
}

int partition(int low,int high)
{
int i;
int last_small;
double pivot;

swap(low,(low+high)/2);
pivot=p[low].x;
last_small=low;
for(i=low+1;i<=high;i++){
if(p[i].x-pivot<eps){
last_small++;
swap(last_small,i);
}
}
swap(low,last_small);
return last_small;
}

void qsort(int i,int j)
{
int pos;
if(i<j){
pos=partition(i,j);
qsort(i,pos-1);
qsort(pos+1,j);
}
}

void merge_sort(int low,int mid,int high)
{
int p1,p2,pointer;
int i;
point temp[10000];

p1=low;
p2=mid+1;
pointer=low;
while(p1<=mid&&p2<=high){
if(p[p1].y-p[p2].y<eps){
temp[pointer].x=p[p1].x;
temp[pointer++].y=p[p1++].y;
}else{
temp[pointer].x=p[p2].x;
temp[pointer++].y=p[p2++].y;
}
}
if(p1<=mid){
temp[pointer].x=p[p1].x;
temp[pointer++].y=p[p1++].y;
}
for(i=low;i<pointer;i++){
p[i].x=temp[i].x;
p[i].y=temp[i].y;
}
}

void init()
{
int i;
for(i=0;i<N;i++)
in>>p[i].x>>p[i].y;
qsort(0,N-1);
}

double dist(int i,int j)
{
return sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
}

double clost_pair(int low,int high)
{
double t,ans1,ans2,ans;
double l;
int index,mid;
point temp[10000];
int i,j,k,s;

if(high-low<3){
for(i=low;i<high;i++)
for(j=i+1;j<=high;j++)
if(p[i].y-p[j].y>eps) swap(i,j);
ans=dist(low,low+1);
for(i=low;i<high;i++)
for(j=i+1;j<=high;j++){
t=dist(i,j);
if(ans-t>eps) ans=t;
}
return ans;
}

mid=(low+high)/2;
l=p[mid].x;
ans1=clost_pair(low,mid);
ans2=clost_pair(mid+1,high);
if(ans1-ans2<eps) ans=ans1;else ans=ans2;
merge_sort(low,mid,high);

index=0;
for(k=low;k<=high;k++){
if((p[k].x-(l-ans)>eps)&&((l+ans)-p[k].x>eps)){
temp[index].x=p[k].x;
temp[index++].y=p[k].y;
}
}
for(k=0;k<index-1;k++)
for(s=k+1;s<index&&s<k+7;s++){
t=sqrt((temp[k].x-temp[s].x)*(temp[k].x-temp[s].x)+(temp[k].y-temp[s].y)*(temp[k].y-temp[s].y));
if(ans-t>eps) ans=t;
}
return ans;
}

void main()
{
double dist;
while(1){
in>>N;
if(!N) break;
init();
if(N==1) printf("INFINITY\n");
else{
dist=clost_pair(0,N-1);
if(10000.0-dist>eps) printf("%.4lf\n",dist);
else printf("INFINITY\n");
}
}
}[/cpp]

[Larry]

Is there trick input on 10245? The acceptance rate seems to be low, and I keep getting WA =(

[rjhadley]

It took me two tries to get Accepted. The only changes between the two versions were:

1) If there is only one point in the data set then output INFINITY instead of garbage. (This is only trick input I can think of).

2) Only use sqrt() when necessary instead of for every distance calculation. (This will speed it up slightly and perhaps avoid some weird floating point rounding errors).

3) Use the reserve() method in my C++ vectors. (For a very minor speed improvement).

So take your pick between #1, #2, and/or a lot of buggy/slow programs being submitted as the cause of the low acceptance rate.

[anak_papua]

why OJ gave my code Output Limit Exceeded?
i didnt see anything that could give me that.
[c]
/*@JUDGE_ID: XXXXX 10245 C*/
#include <stdio.h>
#include <math.h>

#define MAX 10002

typedef struct {
unsigned long x, y;
} tdata;

tdata data[MAX];

int main () {
int jum, i, j;
unsigned long x, y, res, min;
char ch;

while (1) {
scanf ("%d", &jum);
if (jum == 0)
break;

for (i = 0; i < jum; i++)
scanf ("%lu %lu", &data[i].x, &data[i].y);
min = 100000000;
if (jum > 1) {
for (i = 0; i < jum-1; i++)
for (j = i + 1; j < jum; j++) {
x = (data[i].x - data[j].x) * (data[i].x - data[j].x);
y = (data[i].y - data[j].y) * (data[i].y - data[j].y);
res = x + y;
if (res < min)
min = res;
}
}
if (min == 100000000)
printf ("INFINITY\n");
else
printf ("%.4f\n", sqrt (min));
}
return 0;
}
[/c]

[Dominik Michniewski]

Nice hint rhadley :)
I miss in description that for one point I should print INFINITY
But now I'm Accepted this problem and with nice time (less then 0.7 sec)

Thanks again for hint
DM