Give me some tests, please.

"The buses running in one direction should pass the section under repair in the same order they approach it."
Does it means that answer to:

2
1 2
1 1
1
2 2

will be NO?

Why input
6
Zaitseva 21:38.2
Hauswald 21:21.0
Boulygina 22:04.4
Henkel 22:06.1
Wilhelm 21:11.1
Jonsson 22:05.8
Answer:
3
Hauswald
Wilhelm
Zaitseva???????/

Число вершин графа -1

i don't know where could be wrong

#include <cstdio>
#include <string.h>

using namespace std;

struct dat_edge
{
    int last,aim;
}edge[500000];
int temp[300],len_edge,pl[300],next[300],n,m;

void edge_insert(int x,int y)
{
    len_edge++;
    edge[len_edge].aim=y;
    edge[len_edge].last=pl[x];
    pl[x]=len_edge;
}

bool dfs(int root,int ps)
{
    if (temp[root]==ps) return false;
    if (ps==-1) return true;
    temp[root]=ps;
    int p=pl[root];
    bool t;
    while (p!=-1)
    {
        int tmp=edge[p].aim;
        if (temp[tmp]==ps) { p=edge[p].last;continue; }
        if (next[tmp]==-1) t=dfs(tmp,-1); else t=dfs(next[tmp],ps);
        if (t)
        {
            next[root]=edge[p].aim;
            next[edge[p].aim]=root;
            return true;
        }
        p=edge[p].last;
    }
    return false;
}

int main()
{
    //freopen("1099.in","r",stdin);
    //freopen("1099.out","w",stdout);
    int i,j,k,ans,coun,x,y;
    bool ok;
    scanf("%d",&n);
    len_edge=-1;
    for (i = 0;i<=n;i++) pl[i]=-1;
    while (scanf("%d%d",&x,&y)!=EOF)
    {
        edge_insert(x,y);
        edge_insert(y,x);
    }
    coun=0;
    for (i = 0;i<=n;i++) next[i]=-1;
    memset(temp,0,sizeof(temp));
    ans=0;
    for (i = 1;i<=n;i++)
    {
        ok=false;
        if (next[i]==-1) ok=dfs(i,++coun);
        if (ok) ans++;
    }
    printf("%d\n",ans*2);
    for (i = 1;i<=n;i++)
    if (next[i]>i)
    {
        printf("%d %d\n",i,next[i]);
    }
}

Whats test 4 pls ?

i and j can be eqal!

This made my get a WA#6.

Here are some interesting tests:
Input:
1000 4 2 1
1 1
Output:
7

Input:
1000 4 1 1
1 1
Output:
1

Try this test:

Sunday
December 24
17 17

Answer:
1
Monday

Here is my solution. It uses heap(PriorityQueue) and queue(LinkedList). It TLs on 2nd test case. Is there any solution which is asymptotically better than this? What are points to improve?
public class P1126 {
    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

        in.nextToken();
        int m = (int)in.nval;

        PriorityQueue<Integer> heap = new PriorityQueue<Integer>(m, new Comparator<Integer>() {
            public int compare(Integer i, Integer j) {
                return j.compareTo(i);
            }
        });
        LinkedList<Integer>  queue = new LinkedList<Integer>();
        int cur;
        while (true) {
            in.nextToken();
            cur = (int)in.nval;
            if (cur == -1) break;
            heap.add(cur);
            queue.add(cur);
            if (heap.size() == m) {
                System.out.println(heap.peek());
                heap.remove(queue.pollFirst());
            }
        }
    }
}

Hi,

I implemented it using DFA and it is taking around 0.8 seconds. However I can see that some people have done it in 0.03 seconds (that's 27 times faster)? How can it made faster ?

Thanks

Varun

Edited by author 08.03.2010 06:17

my program works fine on my computer. it even gives right answers :) But when i submit it has crash on 20 test. I cannot understant why. Please help!!

import java.io.*;
import java.util.Arrays;
public class Task_1537 {
    public static void main(String[] args) throws Exception{
        StreamTokenizer st = new StreamTokenizer(new InputStreamReader(System.in));
        st.nextToken();
        int k = (int) st.nval;
        st.nextToken();
        int p = (int) st.nval;

        int t[] = new int[k+1];
        t[2] = 1 % p;
        for (int i = 3; i <= k; i++){
            if ((i & 1) == 1)
                t[i] = t[i-1];
            else t[i] = t[i-1] + t[i/2];
            if (t[i] >= p)
                t[i] = t[i] - p;
        }
        //System.out.println(Arrays.toString(t));
        System.out.println(t[k]);
    }

}

there is only one answer or no answer, therefor if the solution exist it's the shortest.
hope can help you.
GOOD LUCK!!!

anyone knows test 6, I have been with this problem for 1 hour

#include<stdio.h>
main()
{
    int N,M,a[1000],i;
    scanf("%i %i",&N,&M);
    for(i=0;i<N;i++)scanf("%i",&a[i]);
    for(i=M;i<M+10;i++)printf("%i",a[i%N]);
}

there is a lot of stuff one has to guess in order to get this problem accepted.

Anyway, watch out for this test:
'a' like '[a-]'

What is in test №2?