// header file
#include <bits/stdc++.h>
// pragma
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
// macros
#define endl "\n"
#define ll long long
#define mp make_pair
#define ins insert
#define lb lower_bound
#define pb push_back
#define ub upper_bound
#define lll __int128
#define fi first
#define se second
using namespace std;
const int lim = 1e5 + 5;
vector<pair<int, int>> edges[lim];
int disc[lim], low[lim];
int t = 0;
vector<int> bridge_id;
// dsu struct
struct disjoint_set {
int h[lim], sz[lim];
disjoint_set() {
memset(h, -1, sizeof(h));
fill(sz, sz + lim, 1);
}
int fh(int nd) {
return h[nd] == -1 ? nd : h[nd] = fh(h[nd]);
}
bool merge(int x, int y) {
x = fh(x), y = fh(y);
if(x != y) {
if(sz[x] < sz[y])
swap(x, y);
sz[x] += sz[y];
h[y] = x;
}
return x != y;
}
};
void dfs(int nd, int pr = -1) {
disc[nd] = low[nd] = t++;
for(auto i : edges[nd]) {
if(i.se == pr) {
continue;
}
else if(disc[i.fi] == -1) {
// undiscovered edge
dfs(i.fi, i.se);
low[nd] = min(low[nd], low[i.fi]);
// check bridge
if(low[nd] < low[i.fi]) {
// case if bridge
bridge_id.pb(i.se);
}
}
else {
// back edge
low[nd] = min(low[nd], disc[i.fi]);
}
}
//cout << "DEBUG " << nd << " " << disc[nd] << " " << low[nd] << endl;
}
// generate subtree count of start/ends and memo par
int st[lim], en[lim], par[lim];
bool vis[lim];
vector<int> bridge_edges[lim];
void bridge_dfs(int nd) {
vis[nd] = 1;
for(auto i : bridge_edges[nd]) {
if(!vis[i]) {
bridge_dfs(i);
st[nd] += st[i];
en[nd] += en[i];
par[i] = nd;
}
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(NULL);
// observe that if an edge is not a bridge, then the edge can face both directions (because it is part of a cycle)
// therefore we only need to mark the bridges. To do this, we can create a bridge tree
int n, m;
cin >> n >> m;
pair<int, pair<int, int>> adj[m + 1];
for(int i = 1; i <= m; ++i) {
int a, b;
cin >> a >> b;
adj[i] = (mp(i, mp(a, b)));
edges[a].pb(mp(b, i));
edges[b].pb(mp(a, i));
}
memset(disc, -1, sizeof(disc));
memset(low, -1, sizeof(low));
dfs(1);
//cout << bridge_id.size() << endl;
//for(auto i : bridge_id)
// cout << i << " ";
//cout << endl;
int p;
cin >> p;
vector<pair<int, int>> v;
for(int i = 1; i <= p; ++i) {
// there exists a road from x -> y
// which means, we need to mark every bridge from x to y as used
int x, y;
cin >> x >> y;
v.pb(mp(x, y));
}
// observe: because there is guaranteed to be a solution. For each bridge that connects A and B,
// we can find the count of start/end of pairings from A and from B.
// If A has more starts, therefore the edge must be A -> B
// else it must be B -> A
// easiest way to create a bridge tree is using DSU
sort(bridge_id.begin(), bridge_id.end());
disjoint_set dsu;
for(int i = 1; i <= m; ++i) {
if(!binary_search(bridge_id.begin(), bridge_id.end(), adj[i].fi)) {
// not a bridge edge
dsu.merge(adj[i].se.fi, adj[i].se.se);
}
}
for(int i = 1; i <= m; ++i) {
// because of some problems, we have to do this in a for loop after the first binary search
// try to think about it
if(binary_search(bridge_id.begin(), bridge_id.end(), adj[i].fi)) {
// is a bridge edge
bridge_edges[dsu.fh(adj[i].se.fi)].pb(dsu.fh(adj[i].se.se));
bridge_edges[dsu.fh(adj[i].se.se)].pb(dsu.fh(adj[i].se.fi));
}
}
for(auto i : v) {
++st[dsu.fh(i.fi)];
++en[dsu.fh(i.se)];
}
bridge_dfs(dsu.fh(1));
// interpret res as result array of direction as one of:
// - 0: B
// - 1: L
// - 2: R
int res[m + 1];
memset(res, -1, sizeof(res));
for(int i = 1; i <= m; ++i) {
if(binary_search(bridge_id.begin(), bridge_id.end(), i)) {
int x = dsu.fh(adj[i].se.fi), y = dsu.fh(adj[i].se.se);
if(par[x] == y) {
if(st[x] > en[x]) {
// x -> y
res[i] = 2;
}
else if(st[x] == en[x]) {
// x <=> y
res[i] = 0;
}
else {
// y -> x
res[i] = 1;
}
}
else {
if(st[y] > en[y]) {
// y -> x
res[i] = 1;
}
else if(st[y] == en[y]) {
// x <=> y
res[i] = 0;
}
else {
// x -> y
res[i] = 2;
}
}
}
else {
res[i] = 0;
}
}
for(int i = 1; i <= m; ++i) {
if(res[i] == 0)
cout << 'B';
else if(res[i] == 1)
cout << 'L';
else if(res[i] == 2)
cout << 'R';
else
assert(false);
}
cout << endl;
return 0;
}
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
2 ms |
7772 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
2 ms |
7772 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
2 ms |
7772 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
