Общий форум"The vertices are listed in the order of traversal" means either in a clockwise or counterclockwise direction. CODE IS HERE; #include "stdio.h" #include "string.h" long a[401][401]; long n,h,t; long i,j,m; int c[500]; int p[500]; long count=0; void bfs(int i) { int k; for(k=1;k<=n;k++) { if(c[k]==0) { if(a[i][k]==1) { a[i][k]=0; a[k][i]=0; c[k]=1; p[k]=i; bfs(k); } } } } void path(int a1,int a2) { int q[500]; int i=0; while(1) { q[i]=a2; i++; a2=p[a2]; if(a2==a1) break; } printf("%d",i+1); printf(" %d",a1); while(i>0) { printf(" %d",q[i-1]); i--; } printf("\n");
}
void main() { for(i=0;i<401;i++) for(j=0;j<401;j++) a[i][j]=0; scanf("%ld %ld",&n,&m); for(i=0;i<m;i++) { scanf("%ld %ld",&h,&t); a[h][t]=1; a[t][h]=1; } for(i=1;i<=n;i++) c[i]=0;
c[1]=1; bfs(1);
for(i=1;i<n;i++) { for(j=i+1;j<=n;j++) if(a[i][j]==1) { count++; } } printf("%ld\n",count); for(i=1;i<n;i++) { for(j=i+1;j<=n;j++) if(a[i][j]==1) { path(i,j); } }
} I have same problem and increased answer's array. got AC Try this 7 8 1 2 2 6 1 6 4 6 4 3 3 5 5 6 3 7 At the moment my way is to look through all the squares of length LEN (using binary search) and to check, whether a hash of a square was met before. It is of O(N*M*log(min{N, M})) complexity. But I get WA#10 again and again. Please, if someone had a problem with this test, share your impressions, 'cause I'm going slightly mad. You have a collisions with hash. (I've got this problem also because second number x was to small and close to first one). Thank you, it really had to do with collisions. Actually, my self-made hashset implementation was fantastically awful. :-) Edited by author 06.07.2012 15:48 Why you have used binary search? How we can say that the sq. matrix 2x2 has less value than 3x3. and 3x3 has less than 4x4. how can we justify? We run a binary search on the length of the square's side. For a fixed length of the square's side, we calculate the hash of all squares with that side length, and look for a pair of squares with the same hash. got WA on 3, why, have any tests? Edited by author 14.03.2024 17:39 Try this: 7 3 6 5 2 6 1 4 4 2 1 1 6 4 1 4 10 2 1 Answer: 1 1 easy max flow Overkill, just dfs is enough) 5 plus 1 plus 1 plus 1 plus 1 plus 1 plus 3 plus 1 plus 1 plus 1 plus 1 plus 1 plus 1 is ... 19!!!!! NINETEEN!!!1111!! FFFFUUUUU~~~~~ Okay, have to read prob statement again. :D You could have used brute force ;) Прошу уточнить. В задании (и тестах) идёт речь о: 1) конечных географических 2D картах. 2) географических картах на сфере. 3) любых абстрактных картах в многомерном \ многосвязном пространстве. 4) другое. Please clarify. The task (and tests) are about: 1) Finite 2D geographical maps. 2) geographical maps on the sphere. 3) favorite abstract maps in a multi-dimensional \ multi-connected space. 4) other. Admins, please look at this issue. I suppose it is mistake (or bug in rating system), that I became 1st place in rating solved same number of tasks as ACRush, but did it only in 2024 year, i.e. later. Suppose that rating must be the same and my place is 2nd. Looks like integer overflow :)) When I resubmitted one single AC solution (for 1394 task) - rating summary has changed again and now I got about 200 points of rating less than 1st place. Edited by author 18.11.2024 22:09 Congrats on solving all the problems! test: 1 10000 0 How do I answer "YES" or "NO"? ooopsss! Zero not allowed! Sorry. Edited by author 29.11.2024 01:34 когда я с длинной арифметикой сделал эту задачу я использовал в качестве основы 10000 и у меня был WA#7 тест потом я сделал снову 10 и у меня был WA#2 тест тогда подумав решил использовать EXTENDED всеравно WA#6 тест кто подскажет как преодолеть 2, 6, 7 тесты Use base = 10 compare small answers with 1009 I compaired it My program gives equal answers in all cases Чёрт его знает Use Java's BigInteger I always use it then long arifmetic is needed. Помоему он подсовывает во 2 тесте N=170 и K=10 Если умножать пллучается 10 с 170 нулями. Никакой тип не выдержит. Надо разбивать число на группы. Только какие? Edited by author 11.06.2008 17:47 174568663359808123517619015636369512622213 064344534689513998708986681415716975079951 752754958954321404709642918328732491072229 658639887075761909680804620151071937764270 946419913401 Edited by author 29.10.2008 11:30 Пока не использовались ostream манипуляторы setfill('0')\setw(9) а использовались только fill('0')\width(9) программа не выводила ведущие нули. AC https://acm.timus.ru/getsubmit.aspx/10809916.cppWA7 https://acm.timus.ru/getsubmit.aspx/10809884.cppЛибо я что-то не понял, либо это ошибка в STL. Надо чтобы админы проверили. P.S. My English is too bad. Понял. Не ошибка STL. Фича. В документации написано (на ангельском языке): Глобально установленная ширина поля имеет право портиться. Edited by author 25.11.2024 04:43If you have an absolutely useless piece of code in your code that is not used anywhere, but it has an error going out of bounds, then this may instead affect the execution of the rest program Round(time) instead of Ceil(time) What should i check if i have WA in test 5? Delete a rectangle if there is another one covering this one All the test cases I've come up with have been solved correctly by this method but no AC :( code: ``` #include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; vector<vector<int>> x(n, vector<int>(3, 0)); for (int i = 0; i < n; i++) { cin >> x[i][0]; cin >> x[i][1]; x[i][2] = i + 1; } sort(x.begin(), x.end()); auto med = (n - 1) / 2; cout << x[med][2] << " " << x[med + 1][2]; } ``` Edited by author 11.11.2024 16:23 Edited by author 11.11.2024 16:23 To answer my own question. The above code will select for the 2 median points. These don't necessarily create a median line. Consider the following test case: 4 0 10 1 1 2 2 3 10 The above alogorithm would select (1, 1) and (2,2), which partitions the plane into a section with 2 points, and a section with 0 points; the above approach is incorrect. Test 3 : input : 5 2 -1 -2 -3 -4 -5 output : 2 4 Test 4 : input : 2 0 1 -5 output : 0 0 Test 5 : input : 15 23 252 33 215 789 1000 123 -125 29 -45 -450 121 -555 -72 120 123 output : 23 113 Test 6 : 0 213 output Impossible Test 7 : input : 8 12 1 2 3 4 5 6 7 8 output : Impossible Test 10 : input : 0 0 output : 0 0 Test 15 : input : 2 12 1000 -1000 output : 12 2012 13 есть? Тест 4 точно не такой. По условию сказано, что x!=0. Я проверял, во всех тестах x!=0. Тест 10 тоже не такой. Edited by author 28.01.2014 07:04 Edited by author 28.01.2014 07:05 правильно подметил Vitaliy. Еще возможен такой вариант : input : 3 2 0 -2 4 output : Impossible Невозможен, "Все координаты, включая x, не равны нулю" You have an error in tests 4 and 10: "Each coordinate, including x, is non-zero and doesn't exceed 1000 in absolute value." 15 isn't it. I have WA15, but on this test my program outputs right answer Send me the answer or I'll kick you! |
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