点联通分量:求割点时已经遍历完一个联通分量,弹栈至改割点即可
题目链接:http://poj.org/problem?id=1523
#include
#include
#include
#include
#include
边双联通分量
题目链接:http://codeforces.com/contest/1000/problem/E
#include
//#define DEBUG
using namespace std;
const int maxn = 3e5 + 5;
int n, m, inde, dcc_num, res;
vector P[maxn], G[maxn];
int low[maxn], dfn[maxn], ID[maxn];
bool vis[maxn];
int dp[maxn];
set> ans;
stack S;
void tarjan(int u, int fa)
{
low[u] = dfn[u] = ++inde;
S.push(u);
for(auto v:P[u])
{
if(v == fa) continue;
if(dfn[v] == -1)
{
tarjan(v, u);
low[u] = min(low[u], low[v]);
if(low[v] > dfn[u])
{
if(u < v)
ans.insert({u, v});
else
ans.insert({v, u});
}
}
else
low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u])
{
dcc_num++;
while(S.top() != u)
{
ID[S.top()] = dcc_num;
S.pop();
}
S.pop();
ID[u] = dcc_num;
}
}
void Get_G()
{
for(int u = 1; u <= n; u++)
{
for(auto v: P[u])
{
if(ID[u] != ID[v])
{
G[ID[u]].push_back(ID[v]);
}
}
}
}
void init()
{
memset(dp, 0, sizeof(dp));
memset(ID, 0, sizeof(ID));
memset(low, -1, sizeof(low));
memset(dfn, -1, sizeof(dfn));
memset(vis, 0, sizeof(vis));
ans.clear();
for(int i = 0; i <= 1005; i++)
P[i].clear(), G[i].clear();
res = dcc_num = inde = 0;
while(!S.empty()) S.pop();
}
void DFS(int u)
{
vis[u] = 1;
for(auto v: G[u])
{
if(vis[v]) continue;
DFS(v);
res = max(res, dp[v] + dp[u] + 1);
dp[u] = max(dp[u], dp[v] + 1);
}
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
init();
int u, v;
for(int i = 0; i < m; i++)
{
cin >> u >> v;
P[u].push_back(v);
P[v].push_back(u);
}
tarjan(1, -1);
Get_G();
DFS(1);
cout << res << endl;
return 0;
}