I did Dijkstra algorithm for S to F, but seems to exceeds time on test #3 ? Anyone got a clue? Tnx in advance

It can be done with only 3 cables with max value of 1.  Got AC

What is the test 4?

In Go you should use `fmt.Scanf("%d %d\n", &a, &b)` instead of `fmt.Scanf("%d %d", &a, &b)`! Using the latter results in reading zeroes.

My max value is 13:
########
#@ $.$.#
#  $.$.#
#$ $.$.#
#. $.$.#
#$ $.$.#
#.   $.#
########

Has anybody found more?

Hint 1: The teams with same scores follow their original order, top to bottom.

Hint 2: You can implement stable sort by yourself.

Hint 3: M is given <=100.

Hint 4: Group all the teams with the same score together in a container.

Solution : Use an array of arrays, with size 101. You can add the input as it comes in the array with the index equal to its score. Output starting from the end, iterating over the array of arrays.

C++ Code:


#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    cin >> n;
    vector<vector<int>> v(101);
    for(int i =0;i<n;i++)
    {
        int a,b;
        cin >>a >> b;
        v[b].push_back(a);
    }
    for(int i =100;i >= 0;i--)
    {
        if((v[i].size()))
        {
            for(auto a:v[i])
            {
                cout << a << " " << i << "\n";
            }
        }
    }
    return 0;
}

I tried with precalculating with map for setting up the index with 1 only rest will be auto 0 but miraculously WA at 4. so i figured out that the author only wants you to solve with his idea only. So good luck.

Missing checks added to the checker program. New tests added to the problem. The time limit increased from 1 sec to 2.5 sec to compensate for increased difficulty.

All solutions were rejudged. 20 out of 45 authors lost the Accepted status for the problem.

Test in you code example like this
4 3
1 5 6
Wrong Answer is 0.
Thrue Answer is 3.
Be confident!!!

/***************************************************************************************************
                         Anzmaul Haque Akash
                    DIU, Daffodil International University
                    E-mail: anzamul15-2297@diu.edu.bd
 Youtube:https://www.youtube.com/channel/UCXvXHUkxyLYKeQkIg_RyMBw
******************************************************************************************************/
#include<bits/stdc++.h>

#define ln     (int)1e6
#define pi     acos(-1.0)
#define pb     push_back
#define pf     push_front 1
#define mp     make_pair
#define F      first
#define S      second
#define INF    1000000000000000000
#define LES    -1000000000000000000
#define intlim 2147483648
#define ll     long long
#define ull    unsigned long long
#define str    string
#define in     insert

#define all(akash)       akash.begin(),akash.end()
#define ms(arr)          memset(arr,0,sizeof(arr))
#define msbooltrue(arr)  memset(arr,true,sizeof(arr))
#define msboolfalse(arr) memset(arr,false,sizeof(arr))
#define msINF(arr)       memset(arr,INF,sizeof(arr))
#define case1(cs)        cout<<"Case "<<cs<<": "
#define case2(cs)        cout<<"Case "<<cs<<":\n"
#define lb               cout<<"\n";

#define vec vector
#define qu  queue
#define li  list
#define pq  priority_queue
#define em  empty

#define fori(i,a,b) for(int i=0;i<b;i++)
#define ford(i,a,b) for(int i=b-1;i>=0;i--)
#define forc(ch);   for(char ch:str)

#define sc1(a)         scanf("%lld",&a)
#define sc2(a,b)       scanf("%lld %lld",&a,&b)
#define sc3(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define sc4(a,b,c,d)   scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sc5(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)

using namespace std;

void solve(ll n){

    ll temp_n=n;

    temp_n--;

    for(ll i=0; i<n-1; i++){
        cout<<"(";
    }
    cout<<"sin(1)+"<<n<<")"; //Segment 1.

    str s="sin(1-sin("; // if n>1 this one is using for the template.


    for(ll i=2;i<=n;i++){ //main loop.

        str temp;
        temp.clear();
        temp=s;


        for(ll x=2; x<i;x++){ //This one creat middle of Every part.

            if(x%2==0){ //' + ' for even ' - ' for odd.
                temp+=to_string(x);
                temp+="+sin(";
            }
            else{
                temp+=to_string(x);
                temp+="-sin(";
            }

        }

        temp+=to_string(i);

        for(ll y=1;y<=i;y++){ //for closing ' ) ' every equation.
            temp+=")";
        }

        cout<<temp; //print each segment.

        if(temp_n > 1){
            cout<<"+"<<temp_n<<")"; //Printing every segment last digit with out 1.
        }
        temp_n--;
    }

    cout<<"+1";lb//This one for end.
}

int main()
{
    ll n;
    sc1(n);

    if(n>1){
        solve(n); //solve function.
    }
    else if(n==1){
        cout<<"sin(1)+1";lb //for 1.
    }

    return 0;
}

AC with max travel dist = 6

My Accepted solution 8202681 for problem 1050 gives incorrect answer on my test (test contains command '\endinputany' in the middle of text).
Please add my test to test package.

     TEST:
There is no "q in this sentence. \par
"Talk child," said the unicorn.

\endinputany
She s\"aid, "\thinspace `Enough!', he said."
\endinput

     INCORRECT ANSWER:
There is no q in this sentence. \par
``Talk child,'' said the unicorn.

\endinputany

     CORRECT ANSWER:
There is no q in this sentence. \par
``Talk child,'' said the unicorn.

\endinputany
She s\"aid, ``\thinspace `Enough!', he said.''
\endinput

New tests added to the problem. Accepted solutions were rejudged. 125 authors lost Accepted status for this problem.

New tests added to the problem. Accepted solutions were rejudged. 144 authors lost Accepted status for this problem.

My AC program prints wrong answer on this test:
9
6 7 1 6
1 8 2 6 5 1
1 2
1 3
2 4
2 5
3 4
4 6
5 6

Correct answer is NO

Hi, Admins.
Please, add these case:
N > 5*10^4
a[i] = 1, for i = 1.. N-1, and
a[N] = 10^9

Q = N
p[i] = i,  x[i] = i

For example:
77888
1 1 1 1 1 1 .... 1 1 77888
77888
1 1
2 2
3 3
4 4
5 5
...
77888 77888

> The student’s home is located at the coordinates (1, 1), and the university is at the coordinates (N, M).
It means, for example, if N == 1 && M == 1 then university have same coordinates as home. Of course we can not use any roads in such case. But according to the sentence:
> There are a total of N streets running from north to south and M streets running from west to east.
there are some roads that can not be used. I think there should be N - 1 streets running from north to south and M - 1 streets running from west to east.

Input Values is not in a range [3, 9]