№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
Usually, topological sort is implemented like
void dfs(int x) {
vis[x] = true;
for(int u = 0; u < n; u++) {
if(!vis[u] && graph[x][u] == 1) {
dfs(u);
}
}
order.push_back(x);
}
And then printed in reverse order But if I implement this way
void dfs(int x) {
order.push_back(x);
vis[x] = true;
for(int u = 0; u < n; u++) {
if(!vis[u] && graph[x][u] == 1) {
dfs(u);
}
}
}
And print in same order.
Can someone provide me a test case where 2nd approach will fail
Название |
---|