답안 #528185

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
528185 2022-02-19T16:09:49 Z cig32 One-Way Streets (CEOI17_oneway) C++17
0 / 100
2 ms 4940 KB
#include "bits/stdc++.h"
using namespace std;
mt19937_64 rng((int)std::chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
#define int long long
 
int rnd(int x, int y) {
  int u = uniform_int_distribution<int>(x, y)(rng); return u;
}
vector<int> adj[MAXN];
int tin[MAXN], tout[MAXN], low[MAXN], vis[MAXN], tim = 0;
map<pair<int, int>, bool> is_bridge;
map<pair<int, int>, int> o; // appear count
void dfs(int node, int prv) {
  tin[node] = ++tim;
  int mi = 1e9;
  vis[node] = 1;
  for(int x: adj[node]) {
    if(!vis[x]) {
      dfs(x, node);
      mi = min(mi, low[x]);
      //cout << "tree edge: " << node << " -> " << x << "\n";
      if(low[x] > tin[node] && o[{x, node}] == 1) {
        //cout << "bridge found: " << x << " " << node << "\n";
        is_bridge[{x, node}] = is_bridge[{node, x}] = 1;
      }
    }
  }
  tout[node] = ++tim;
  //Calculating low[node]
  low[node] = tin[node];
  for(int x: adj[node]) {
    if(tin[x] < tin[node] && tout[x] == 0 && tin[x] > 0 && x != prv) {
      //cout << x << " is an ancestor of " << node << "\n";
      low[node] = min(low[node], tin[x]);
    }
  }
  low[node] = min(low[node], mi);
  //cout << node << ": " << tin[node] << " " << tout[node] << " " << low[node] <<" \n";
}
int dsu[MAXN];
int set_of(int u) {
  if(dsu[u] == u) return u;
  return dsu[u] = set_of(dsu[u]);
}
void union_(int u, int v) {
  dsu[set_of(u)] = set_of(v);
}
vector<int> adj2[MAXN];
map<pair<int, int>, int> ord;
map<pair<int, int>, int> rel;
bool dfs2(int node, int prv, int tar) {
  bool good = (node == tar);
  if(!good) {
    for(int x: adj2[node]) {
      if(x != prv) {
        if(dfs2(x, node, tar)) {
          good = 1;
        }
      }
    }
  }
  if(good) rel[{prv, node}] = rel[{node, prv}] = prv;
  return good;
}
void solve(int tc) {
  int n, m;
  cin >> n >> m;
  for(int i=1; i<=n; i++) {
    dsu[i] = i;
  }
  pair<int, int> edges[m+1];
  for(int i=1; i<=m; i++) {
    int a, b;
    cin >> a >> b;
    edges[i] = {a, b};
    o[{a, b}]++; o[{b, a}]++;
    if(a == b) continue;
    if(o[{a, b}] > 1) continue;
    ord[edges[i]] = a;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, -1);
  for(int i=1; i<=n; i++) {
    for(int x: adj[i]) {
      if(!is_bridge[{i, x}] || o[{i, x}] > 1) {
        if(set_of(i) != set_of(x)) {
          union_(i, x);
        }
      }
    }
  }
  map<pair<int, int>, bool> alr;
  for(int i=1; i<=n; i++) {
    for(int x: adj[i]) {
      if(set_of(i) != set_of(x) && !alr[{set_of(i), set_of(x)}]) {
        alr[{set_of(i), set_of(x)}] = alr[{set_of(x), set_of(i)}] = 1;
        adj2[set_of(i)].push_back(set_of(x));
        adj2[set_of(x)].push_back(set_of(i));
      }
    }
  }
  //adj2 is the bridge-block tree of the graph
  int p; cin >> p;
  for(int i=0; i<p; i++) {
    int x, y;
    cin >> x >> y;
    x = set_of(x), y = set_of(y);
    if(x == y) continue;
    dfs2(x, -1, y);
  }
  for(int i=1; i<=m; i++) {
    if(edges[i].first == edges[i].second) {
      cout << "B"; continue;
    }
    else if(o[edges[i]] > 1) {
      cout << "B"; continue;
    }
    else if(!is_bridge[edges[i]]) {
      cout << "B"; continue;
    }
    int r = rel[{set_of(edges[i].first), set_of(edges[i].second)}];
    int q = set_of(ord[edges[i]]);
    if(r > 0) cout << (r == q ? "R" : "L");
    else cout << "B";
  }
  cout << "\n";
}

int32_t main(){
  ios::sync_with_stdio(0); cin.tie(0);
  int t = 1;// cin >> t;
  for(int i=1; i<=t; i++) solve(i);
}  
/*

5 7
1 2
1 2
4 3
2 3
1 3
5 1
1 5
2
5 3
4 1

*/
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 4940 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 4940 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 4940 KB Output isn't correct
2 Halted 0 ms 0 KB -