got AC but shouldn't have. i used map to keep mark upto what edges outgoing from a certain node we have already dp'ed. and return in log(n). but when i have to dp more then dp'ed and entered into map for that node. however, this gave wa 17. then i dp'ed some more node although already dp'ed. if this 'some more' is less than 9 then i get wa on (18~23) tests. but when it is >=9 i get AC. i dunno why my soln was WA, it seemed perfect. but even more surprising when i get AC like this. Portion of my code:-

#define WHATTHEHECK 9
...

map<int,double> optimize[MAX];
...

int lim1=arr[edgeno].dtime+arr[edgeno].dur,lim2=lim1+arr[edgeno].delay;
    int s1=v[to].size(),s2=s1;
    int lo=0,hi=s1-1,mid;
    while(lo<=hi){
        mid=(lo+hi)/2;
        if(lim1<=arr[v[to][mid]].dtime){
            hi=mid-1,s1=mid;
        }else lo=mid+1;
    }
    lo=0,hi=s2-1;
    while(lo<=hi){
        mid=(lo+hi)/2;
        if(lim2<=arr[v[to][mid]].dtime){
            hi=mid-1,s2=mid;
        }else lo=mid+1;
    }
    int i=-1;
    if(optimize[to].size()==0) i=v[to].size()-1;
    else i=min((*optimize[to].begin()).first+WHATTHEHECK,v[to].size()-1);
    if(i>=s1){
        for(;i>=s1;--i){
            optimize[to][i]=ret1=min(ret1,dp(v[to][i]));
        }
    }
    if(s1<v[to].size()) ret1=optimize[to][s1];
    if(s2<v[to].size()) ret2=optimize[to][s2];

ps: i would be really glad if someone explain me what i did wrong and what's going on here
:)

I got WA17 when I wrongly (overwrite old mean) insert into segment tree flights with same departure airport and same departure time.

Please use english. I translated your question with google translate and it said "How do you know when it has completed writing a message?".
Well, it depends on the programming language you are using.

Pascal: while not(eof(input)) (not sure if it was input)
C: you usualy use while(!feof(stdin)) http://www.cplusplus.com/reference/cstdio/feof/
   This doesn't seem to work (for me at least).
C++: I think it's !cin.eof() (I've never used this one).

on pascal

while not eoln do
 begin
  read(g_string);
  ......
  ......
 end;

on java:
 Scanner scan = new Scanner(System.in);
while (scan.hasNext()){
  s = scan.nextLine();
}

on python
message = sys.stdin.readlines()
and then ctrl+D to stop input

Use bubble sort and THEN assert that the balls are the same color in (n / 2) and (n / 2 + 1)

My AC program writes following answers:

80 60 >> 199
60 80 >> 200
60 60 >> 160

BTW I solved this problem without any boring special cases consideration, using only stupid DP[101][101]...

80 60     199
60 80     200
60 60     160

Its quite easy to do.
Basically, there are 2 "worst" cases:
1). We take 40 right (equals 40*2=80 secs), then throw all the rest of the right ones (2*(b-40) secs), and them take the 40 left (40 secs). The time is 80+2*(b-40)+40=120+2*(b-40).
2). We take 39 left (78 secs), take 40 left ones (40 secs), throw all the other left ones away (2*(a-40)), and take the last right one (1 sec). The time is 78+40+2*(a-40)+1=119+2*(a-40).

The answer is the maximum of these two.

Not sure, but it could be because of this t[4*MAXN] instead of t[400000]

Yeah, he correctly thought that people will forgive him, because it is charity

Use this site to visualise the graph using matrix and firstly change enumeration from 1,2,3... to 0,1,2... :
https://graphonline.ru/#

Edited by author 12.05.2023 13:55

Edited by author 12.05.2023 13:55