LightOj 1291 (Real Life Traffic)
Idea⌗
Articulation Bridge
- Find the articulation bridges in the graph.
- If you remove the bridges, you will get several connected components.
- Think carefully, there are 2 types of components.
- The 1st one is the connected component which had more than one bridge, if you cut one bridge, it still can connect with other components through other bridge.
- The 2nd one is the component which had only one bridge, if you cut the bridge, it cannot connect with other components.
- You may think that the answer is the number of 2nd type bridge, as they need another one path to connect with other components.
- Think carefully, you may still minimize the number of edges required.
- If you connect one 2nd type component to another 2nd type component, then it requires one edge rather than two edges.
/** Which of the favors of your Lord will you deny ? **/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define MP make_pair
#define F first
#define S second
#define INF INT_MAX
#define ALL(x) (x).begin(), (x).end()
#define DBG(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << endl
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class TIn>
using indexed_set = tree<
TIn, null_type, less<TIn>,
rb_tree_tag, tree_order_statistics_node_update>;
/*
PBDS
-------------------------------------------------
1) insert(value)
2) erase(value)
3) order_of_key(value) // 0 based indexing
4) *find_by_order(position) // 0 based indexing
*/
inline void optimizeIO()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const int nmax = 1e4+7;
const LL LINF = 1e17;
string to_str(LL x)
{
stringstream ss;
ss<<x;
return ss.str();
}
//bool cmp(const PII &A,const PII &B)
//{
//
//}
vector<int>adj[nmax];
set<int>newGraph[nmax];
vector<bool>visited;
vector<int>SCCMap;
vector<int> discov; /** Discovery time in DFS **/
vector<int> low; /** min(all discovery time of subtree of a vertex u including the back-edge ancestors) **/
vector<PII> articulationBridge;
int timer;
int scc = 0;
void initialize()
{
timer = 0;
visited.assign(nmax,false);
SCCMap.assign(nmax,-1);
discov.assign(nmax,-1);
low.assign(nmax,-1);
articulationBridge.clear();
for(int i=0; i<nmax; i++)
adj[i].clear() , newGraph[i].clear();
}
void dfs(int v,int p)
{
visited[v] = true;
discov[v] = low[v] = timer++;
int child = 0;
for(int next:adj[v])
{
child++;
if(next==p)
continue;
if(visited[next])
low[v] = min(low[v],discov[next]);
else
{
dfs(next,v);
low[v] = min(low[v],low[next]);
if(discov[v]<low[next])
{
articulationBridge.push_back({v,next});
newGraph[v].erase(next);
newGraph[next].erase(v);
}
}
}
}
void scc_dfs(int u)
{
visited[u] = true;
SCCMap[u] = scc;
for(int next:newGraph[u])
{
if(!visited[next])
scc_dfs(next);
}
}
int main()
{
//freopen("out.txt","w",stdout);
optimizeIO();
int tc;
cin>>tc;
for(int q=1;q<=tc;q++)
{
initialize();
int n,m;
cin>>n>>m;
for(int i=1; i<=m; i++)
{
int a,b;
cin>>a>>b;
adj[a].push_back(b);
adj[b].push_back(a);
newGraph[a].insert(b);
newGraph[b].insert(a);
}
for(int i=0; i<n; i++)
{
if(!visited[i])
dfs(i,-1);
}
visited.assign(nmax,false);
scc = 0;
for(int i=0; i<n; i++)
{
if(!visited[i])
{
scc_dfs(i);
scc++;
}
}
for(auto bridge:articulationBridge)
{
numBridgesConToComp[SCCMap[bridge.F]]++;
numBridgesConToComp[SCCMap[bridge.S]]++;
}
int cc = 0;
for(int i=0;i<scc;i++)
{
if(numBridgesConToComp[i]==1)
cc++;
}
int ans = (cc+1)/2;
cout<<"Case "<<q<<": ";
cout<<ans<<endl;
}
return 0;
}
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