/**
* Notes during contest.
*
* ------ A ------
* Looks like that it's about strongly-connected components, cycles or bridges. I
* think we can detect all cycles and merge all those nodes into one. The resulting
* graph would be a tree, so we can then use a difference array to find each edge's
* direction. Note that any edge in a cycle is B, and any edge not in cycles is a
* "bridge".
* !IMPORTANT! (Reminded by tinjyu during contest) !IMPORTANT!
* Graph may not be CONNECTED.
*
* ------ B ------
* We iterate through all cost k. We can take at most k bets. If we can guarantee
* a profit of p, min(#A_is_the_outcome, #B_is_the_outcome) >= p + k. Binary search
* is ok.
* For each k, we wanna find the max achievable for
* min(#A_is_outcome, #B_is_outcome) - k.
*
* ------ C ------
*
* Time Complexity 1: O((n + m) * p)
* Time Complexity 2: O(n * log(n * VAL_MAX) * log(n))
* Time Complexity 3: O()
* Implementation 0.7 (Solves [S1-2])
*/
#include <bits/stdc++.h>
typedef int64_t int_t;
typedef std::vector<int_t> vec;
const int_t INF = 0x3f3f3f3f3f3f;
// a DSU impl that allows assigning a value to each set
struct DSU {
vec set, rank;
int_t components;
DSU(int_t n) {
set.resize(n);
rank.resize(n);
for (int_t k = 0; k < n; k++)
set[k] = k, rank[k] = 1;
components = n;
}
inline int_t find(int_t k) {
if (set[k] == k)
return k;
return set[k] = find(set[k]);
}
bool merge(int_t a, int_t b) { // merge b into a (keep a's value)
a = find(a), b = find(b);
if (a == b)
return false;
if (rank[b] > rank[a])
std::swap(a, b);
set[b] = a, rank[a] += rank[b], components--;
return true;
}
};
struct edge_t {
int_t a, b;
};
struct neighb_t {
int_t neighb, edge_idx;
};
typedef std::vector<neighb_t> adj_list_t;
void dfs1(int_t k, int_t parent, const std::vector<adj_list_t>& graph,
std::vector<bool>& tree_edge, vec& depth, std::vector<bool>& visited) {
visited[k] = true;
if (parent != -1)
depth[k] = depth[parent] + 1;
else
depth[k] = 0;
for (const neighb_t& child : graph[k]) {
if (child.neighb == parent || visited[child.neighb])
continue;
tree_edge[child.edge_idx] = true;
dfs1(child.neighb, k, graph, tree_edge, depth, visited);
}
}
void dfs2(int_t k, int_t parent, const std::vector<adj_list_t>& graph,
const std::vector<bool>& tree_edge, vec& across, const vec& depth,
std::vector<bool>& is_bridge) {
across[k] = 0;
int_t backward_up = 0, backward_down = 0;
for (const neighb_t& child : graph[k]) {
if (tree_edge[child.edge_idx] && child.neighb != parent) {
dfs2(child.neighb, k, graph, tree_edge, across, depth, is_bridge);
across[k] += across[child.neighb];
} else {
if (depth[child.neighb] > depth[k])
backward_down++;
else if (depth[child.neighb] < depth[k])
backward_up++;
}
}
if (parent != -1) // the tree edge linking to parent
backward_up--;
across[k] += backward_up - backward_down;
assert(across[k] >= 0);
if (parent != -1 && across[k] == 0) {
for (const neighb_t& child : graph[k]) {
if (child.neighb == parent)
is_bridge[child.edge_idx] = true;
}
}
}
// Find bridges in the graph with a DFS tree
// I learnt this from a tutorial on codeforces (about DFS trees)
std::vector<bool> find_bridge(int_t n, int_t m,
const std::vector<edge_t>& original_graph) {
std::vector<adj_list_t> graph(n);
for (int_t k = 0; k < m; k++) {
int_t a = original_graph[k].a, b = original_graph[k].b;
graph[a].push_back(neighb_t{b, k});
graph[b].push_back(neighb_t{a, k});
}
vec depth(n), across(n);
std::vector<bool> tree_edge(m, false);
std::vector<bool> visited(n, false);
std::vector<bool> is_bridge(m, false);
for (int_t k = 0; k < n; k++) {
if (!visited[k]) {
dfs1(k, -1, graph, tree_edge, depth, visited);
dfs2(k, -1, graph, tree_edge, across, depth, is_bridge);
}
}
return is_bridge;
}
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
int_t n, m;
std::cin >> n >> m;
std::vector<edge_t> original_graph(m);
for (int_t k = 0; k < m; k++) {
std::cin >> original_graph[k].a >> original_graph[k].b;
original_graph[k].a--, original_graph[k].b--;
}
std::vector<bool> is_bridge = find_bridge(n, m, original_graph);
std::cerr << "[debug] is_bridge[]: ";
for (int_t k = 0; k < m; k++)
std::cerr << is_bridge[k];
std::cerr << std::endl;
DSU compress(n); // used to compress the nodes
for (int_t k = 0; k < m; k++) {
if (!is_bridge[k])
compress.merge(original_graph[k].a, original_graph[k].b);
}
vec map(n, -1); // mapping an original node into one in our new graph
// std::cerr << "[debug] map[]: ";
for (int_t k = 0, new_graph_top = 0; k < n; k++) {
if (map[compress.find(k)] == -1)
map[compress.find(k)] = new_graph_top++;
map[k] = map[compress.find(k)];
// std::cerr << map[k] << ' ';
}
// std::cerr << std::endl;
std::vector<adj_list_t> graph(compress.components);
for (int_t k = 0; k < m; k++) {
int_t a = map[original_graph[k].a], b = map[original_graph[k].b];
if (a != b) {
graph[a].push_back(neighb_t{b, k});
graph[b].push_back(neighb_t{a, k});
}
}
n = compress.components;
int_t pairs;
std::cin >> pairs;
vec start(m, -1); // for each edge, if it's a -> b, start[] = a.
for (int_t k = 0; k < pairs; k++) {
int_t src, dst;
std::cin >> src >> dst;
src = map[src - 1], dst = map[dst - 1];
if (src == dst)
continue;
// TODO: improve BFS
std::queue<int_t> bfs_queue;
std::vector<int_t> edge_num(n, -1), prev(n, -1);
edge_num[src] = INF; // doesn't exist. Anything not -1 would be good
bfs_queue.emplace(src);
while (!bfs_queue.empty()) {
int_t t = bfs_queue.front();
bfs_queue.pop();
if (t == dst)
break;
for (const neighb_t& e : graph[t]) {
if (edge_num[e.neighb] != -1)
continue;
edge_num[e.neighb] = e.edge_idx, prev[e.neighb] = t;
bfs_queue.push(e.neighb);
}
}
for (int_t pt = dst; pt != src; pt = prev[pt]) {
assert(edge_num[pt] != -1);
start[edge_num[pt]] = prev[pt];
}
}
std::string ans(m, '@');
for (int_t k = 0; k < m; k++) {
if (is_bridge[k]) {
if (start[k] != -1) {
if (map[original_graph[k].a] == start[k])
ans[k] = 'R';
else
ans[k] = 'L';
} else {
ans[k] = 'B';
}
} else {
ans[k] = 'B';
}
}
std::cout << ans << '\n';
}
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Correct |
0 ms |
212 KB |
Output is correct |
| 3 |
Correct |
2 ms |
468 KB |
Output is correct |
| 4 |
Correct |
3 ms |
468 KB |
Output is correct |
| 5 |
Correct |
3 ms |
468 KB |
Output is correct |
| 6 |
Correct |
2 ms |
468 KB |
Output is correct |
| 7 |
Correct |
3 ms |
516 KB |
Output is correct |
| 8 |
Correct |
3 ms |
468 KB |
Output is correct |
| 9 |
Correct |
2 ms |
340 KB |
Output is correct |
| 10 |
Correct |
2 ms |
340 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Correct |
0 ms |
212 KB |
Output is correct |
| 3 |
Correct |
2 ms |
468 KB |
Output is correct |
| 4 |
Correct |
3 ms |
468 KB |
Output is correct |
| 5 |
Correct |
3 ms |
468 KB |
Output is correct |
| 6 |
Correct |
2 ms |
468 KB |
Output is correct |
| 7 |
Correct |
3 ms |
516 KB |
Output is correct |
| 8 |
Correct |
3 ms |
468 KB |
Output is correct |
| 9 |
Correct |
2 ms |
340 KB |
Output is correct |
| 10 |
Correct |
2 ms |
340 KB |
Output is correct |
| 11 |
Correct |
167 ms |
11052 KB |
Output is correct |
| 12 |
Correct |
200 ms |
12516 KB |
Output is correct |
| 13 |
Correct |
169 ms |
14380 KB |
Output is correct |
| 14 |
Correct |
185 ms |
16000 KB |
Output is correct |
| 15 |
Correct |
216 ms |
16488 KB |
Output is correct |
| 16 |
Correct |
217 ms |
15448 KB |
Output is correct |
| 17 |
Correct |
649 ms |
18324 KB |
Output is correct |
| 18 |
Correct |
793 ms |
15560 KB |
Output is correct |
| 19 |
Correct |
529 ms |
20428 KB |
Output is correct |
| 20 |
Correct |
181 ms |
11980 KB |
Output is correct |
| 21 |
Correct |
153 ms |
11460 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Correct |
0 ms |
212 KB |
Output is correct |
| 3 |
Correct |
2 ms |
468 KB |
Output is correct |
| 4 |
Correct |
3 ms |
468 KB |
Output is correct |
| 5 |
Correct |
3 ms |
468 KB |
Output is correct |
| 6 |
Correct |
2 ms |
468 KB |
Output is correct |
| 7 |
Correct |
3 ms |
516 KB |
Output is correct |
| 8 |
Correct |
3 ms |
468 KB |
Output is correct |
| 9 |
Correct |
2 ms |
340 KB |
Output is correct |
| 10 |
Correct |
2 ms |
340 KB |
Output is correct |
| 11 |
Correct |
167 ms |
11052 KB |
Output is correct |
| 12 |
Correct |
200 ms |
12516 KB |
Output is correct |
| 13 |
Correct |
169 ms |
14380 KB |
Output is correct |
| 14 |
Correct |
185 ms |
16000 KB |
Output is correct |
| 15 |
Correct |
216 ms |
16488 KB |
Output is correct |
| 16 |
Correct |
217 ms |
15448 KB |
Output is correct |
| 17 |
Correct |
649 ms |
18324 KB |
Output is correct |
| 18 |
Correct |
793 ms |
15560 KB |
Output is correct |
| 19 |
Correct |
529 ms |
20428 KB |
Output is correct |
| 20 |
Correct |
181 ms |
11980 KB |
Output is correct |
| 21 |
Correct |
153 ms |
11460 KB |
Output is correct |
| 22 |
Execution timed out |
3049 ms |
18444 KB |
Time limit exceeded |
| 23 |
Halted |
0 ms |
0 KB |
- |
