OJ Board

I am getting RTE, help me out by giving me valid input
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
#include<string.h>
#include<ctype.h>
#define T 1000
#define Z 1000000
typedef struct node *link;
struct node{int n;
int team;
link prev;
link next;
};
link NEW(int n,int team,link prev ...

ANS: keep an updated link to the last element of each group thus reducing ENQUEUE to O(1)

My algorithm times-out during it's run. I am asking for help optimizing my approach. I keep two doubly-linked lists (1) a freelist (2) the team queue. I start by adding every element into the freelist and remove each as they are ENQUEUEd into the team queue. Searching the freelist and adding into to ...

1,2,3 hold the same opinions as 4,5 and 6 respectively this makes 3 opinions in total not 4. Is there something I am missing? I have read the problem statement many times

I have used all the test data in the forum but I still get wrong answer; here is my program.

#include<iostream>
#include<functional>
#include<algorithm>
#include<iterator>
#include<string>
#include<sstream>
#include<climits>
#include<cassert>
using namespace std;
#define LIM 2147483647
int main ...

You can choose any shortest path algorithm so long as you compute the predecessor matrix/array. Then remove the shortest path and run the algorithm once more; you will end up with the alternate shortest path. If you find a way of computing all (same length) shortest paths with a single pass of the ...

I also took care of:
4 4
0 1 10
1 2 5
1 2 0
2 3 15

Vertices 1 & 2 are the same; the edge of length 5 from 1 to 2 is a selfloop. Thus the path from 0 to 3 is even.

I am looking for more critical input

Do not treat the doubles as doubles, read them in as strings. Ohh dear, have I said too much

You will continue to get WA until you have the right number of dashes above every recond. Copy the dash line from the problem's output (there are 39 if I am not mistaken) paste it on your code. I had my program ACed after doing so; give this a try.

int main()
{
int n;
cin.ignore(4); // '1' + '\n' + '2' + '\n' = 4 chars
cin>>n;
cout<<n<<endl; // output: 3
}

There is an error in problem 10183 (How many Fibs?) ..... F[2] is defined as 2 instead of 1; you must however use the information as given in order to get an AC.

Sorry, was just following the [bad] example of others on the board. I will have to do the work myself. On a related note, I just got a copy of the book "The Practice of Programming" thought I might learn good programming habits from the get-go. From the look of things many on the forum should ...

Just looking for input that could break my program, source is for ur experimentation pleasure

I have checked previous posts and know for sure that my solution is correct and most likely what first comes to mind when dealing with this problem. However, I still can't get mine excepted. If u have any ideas I would love to hear from u.

Thanks, Robert

#include <stdio.h>
#include <stdlib.h ...