Solving 10385 - Duathlon (Ternary search)
[andmenj-acm.git] / 10199 - Tourist guide / 10199.2.cpp
bloba36185235fb2c16e8d3d982f3473e719e16cfc7a
1 #include <vector>
2 #include <set>
3 #include <map>
4 #include <algorithm>
5 #include <iostream>
6 #include <iterator>
8 using namespace std;
10 typedef string node;
11 typedef map<node, vector<node> > graph;
12 typedef char color;
14 const color WHITE = 0, GRAY = 1, BLACK = 2;
16 graph g;
17 map<node, color> colors;
18 map<node, int> d, low;
20 set<node> cameras;
22 int timeCount;
24 void dfs(node v, bool isRoot = true){
25 colors[v] = GRAY;
26 d[v] = low[v] = ++timeCount;
27 vector<node> neighbors = g[v];
28 int count = 0;
29 for (int i=0; i<neighbors.size(); ++i){
30 if (colors[neighbors[i]] == WHITE){ // (v, neighbors[i]) is a tree edge
31 dfs(neighbors[i], false);
32 if (!isRoot && low[neighbors[i]] >= d[v]){
33 cameras.insert(v);
35 low[v] = min(low[v], low[neighbors[i]]);
36 ++count;
37 }else{ // (v, neighbors[i]) is a back edge
38 low[v] = min(low[v], d[neighbors[i]]);
41 if (isRoot && count > 1){ //Is root and has two neighbors in the DFS-tree
42 cameras.insert(v);
44 colors[v] = BLACK;
47 int main(){
48 int n;
49 int map = 1;
50 while (cin >> n && n > 0){
51 if (map > 1) cout << endl;
52 g.clear();
53 colors.clear();
54 d.clear();
55 low.clear();
56 timeCount = 0;
57 while (n--){
58 node v;
59 cin >> v;
60 colors[v] = WHITE;
61 g[v] = vector<node>();
64 cin >> n;
65 while (n--){
66 node v,u;
67 cin >> v >> u;
68 g[v].push_back(u);
69 g[u].push_back(v);
72 cameras.clear();
74 for (graph::iterator i = g.begin(); i != g.end(); ++i){
75 if (colors[(*i).first] == WHITE){
76 dfs((*i).first);
80 cout << "City map #"<<map<<": " << cameras.size() << " camera(s) found" << endl;
81 copy(cameras.begin(), cameras.end(), ostream_iterator<node>(cout,"\n"));
82 ++map;
84 return 0;