I get a Wrong answer on test case 6 on this problem. Would it be possible (and legal, of course) for me to see the test case and/or a similar case that demonstrates why my solution is incorrect so I can test it locally?

please, give me any test for WA #7

Edited by author 17.10.2010 20:13

:)
Thank you!

Can someone write the sequence of stone that lead to player 2 lose plz ?

Please answer,waht is the right answer for this text`
The Machinegunners played away game, scored 0 goals, and conceded 0 goals.
?

#include <iostream>
#include <math.h>
using namespace std;

float tar[100000][2];
float slad[10000][2];
float ooo[100000][2];

int prov (float a[100000][2], int n) {
    int i=0; int j=0;
    for (i=0; i<n; i++)
    for (j=i+1; j<n; j++)
    if (a[i][0]==a[j][0] || a[i][1]==a[i][1]) return 0;
    return 1; }

void obch (float x, float y, float tarr[100000][2], int mm, float oooo[100000][2]) {
    int ii=0;
    for (ii=0; ii<mm; ii++) {
        oooo[ii][0]=sqrt(((x-tarr[ii][0])*(x-tarr[ii][0]))+((y-tarr[ii][1])*(y-tarr[ii][1])));
        oooo[ii][1]=ii+1; } }

float min_n (float oooo[100000][2], int mm) {
    float nomm=oooo[0][1]; float znn=oooo[0][0]; int ii=0;
    for (ii=0; ii<mm; ii++) {
        if (oooo[ii][0]<znn) { nomm=oooo[ii][1]; znn=oooo[ii][0]; } }
    return nomm; }

int main () {
    int i=0; int m; int n;
    float min;
    cin>>m;
    for (i=0; i<m; i++) {
        cin>>tar[i][0];
        cin>>tar[i][1]; }
    cin>>n;
    if(m<=n || m<=0) return 0;
    for (i=0; i<n; i++) {
        cin>>slad[i][0];
        cin>>slad[i][1]; }
    if (!prov(tar,m) && !prov(slad,n)) return 0;
    for (i=0; i<n; i++) {
        obch (slad[i][0], slad[i][1], tar, m, ooo);
        min=min_n(ooo,m);
        for (float j=min-1; j<m; j++)
        if (ooo[(int)(min-1)][0]==ooo[(int)(j)][0]) cout<<ooo[(int)(j)][1]<<' ';
        cout<<endl; }
    system ("pause");
    return 0; }

is it possible to find biparite matching for n or n logn?
if we divide graph into 2 parts  - maximal matching is the answer. But biparite matching founding time is NM(max-flow) so N*N(because it's tree).

Edited by author 09.07.2011 10:54

use priority_queue, it eats less memory

import java.util.Arrays;
import java.util.Scanner;


public class Example2 {

    private static int[] col;
    private static int count = 1;
    private static int superCount = 0;

    public static void main(String[] args) {
        try {
            Scanner scanner = new Scanner(System.in);
            int a = scanner.nextInt();
            if (a>=0 && a<=150){
                col = new int[a];
                for (int i = 0; i < a; i++) {
                    col[i]=scanner.nextInt();
                    if(col[i]<600 || col[i]>700){
                        //System.out.print("0");
                        throw new Exception();
                    }
                }
                Arrays.sort(col);
                for (int i = 1; i < col.length; i++) {
                    if (col[i]==col[i-1]) {
                        count ++;
                        if (count==4) {
                            superCount ++;
                            count=1;
                        }
                    }else count=1;
                }
                System.out.print(superCount);
            }else if (a>=0){System.out.print("0");}
        } catch (Exception e) {}
    }

}

program ural_1001;
var
t,n,i,j:longint;
a:array [0..1000000] of real;
begin
while not eoln do
 begin
  inc(t);
  read(a[t]);
  a[t]:=sqrt(a[t]);
 end;
for i:=t downto 1 do
 writeln(a[i]:0:4)
 end.

bfs

Edited by author 06.07.2011 10:45

import java.util.Scanner;


public class Example {

    private static int n;
    private static String[] penguin;
    private static int[] myCount = {1, 1, 1};

    public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            n = scanner.nextInt();
            if(n>=1 && n<=1000){
                scanner = new Scanner(System.in);
                penguin = new String[n];
                for (int i = 0; i < n; i++) {
                    penguin[i] = scanner.nextLine();
                    if (penguin[i].equals("Emperor Penguin")) {
                        myCount[0]++;
                    }else if (penguin[i].equals("Little Penguin")) {
                        myCount[1]++;
                    }else if (penguin[i].equals("Macaroni Penguin")) {
                        myCount[2]++;
                    }
                }
                if(myCount[0]>myCount[1]){
                    if(myCount[0]>myCount[2]){
                        System.out.println("Emperor Penguin");
                    }else{
                        System.out.println("Macaroni Penguin");
                    }
                }else if(myCount[1]>myCount[2]){
                    System.out.println("Little Penguin");
                }else{
                    System.out.println("Macaroni Penguin");
                }

            }
    }

}

i have write NlogN algo -  TLE  21
then i rewrite it in N*sqrt(N) - TLE 21
maybe something wrong with this test?

3
2 1
2 3
2 4
This sample correct?

Edited by author 14.07.2010 13:02

New test has been added and my program had WA on test 103. Can someone give me this test?