zju1119 SPF

终于写好无向图找割点了，笔记一下

割点的充分必要条件是
1. 对于dfs搜索树的根，有至少两个子树
2. 对于非dfs树的根，有至少一个子树

p.s.其实这两个条件是一样的，因为对于非根节点，有一条边连到父节点，所以把它看成是树根的时候，就有至少两个子树了

[code="c++"]
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
using namespace std;

vector<vector<int> > g;
set<int> node;

vector<int> dfn, low, sub;
int cnt;
void dfs(int u)
{
dfn[u] = low[u] = cnt++;
for (int i = 0; i < g[u].size(); ++i)
{
int v = g[u][i];
if (dfn[v] == -1) {
dfs(v);
low[u] = min(low[u], low[v]);
if (dfn[u] <= low[v])
++sub[u];
} else {
low[u] = min(low[u], dfn[v]);
}
}
}

int main()
{
int tcs = 1, u, v;
while (true)
{
g.clear();
node.clear();
bool read = false;
while (scanf("%d", &u), u != 0)
{
read = true;
scanf("%d", &v);
int t = max(u, v) + 1;
t = max((int)g.size(), t);
g.resize(t);
g[u].push_back(v);
g[v].push_back(u);
node.insert(u);
node.insert(v);
}
if (!read)
break;

dfn.assign(g.size(), -1);
low.assign(g.size(), 0);
sub.assign(g.size(), 0);
cnt = 0;
int root = *node.begin();
dfs(root);
--sub;
if (tcs > 1)
printf("\n");
printf("Network #%d\n", tcs++);
bool has = false;
for (int i = 0; i < g.size(); ++i)
if (sub[i] >= 1)
{
has = true;
printf(" SPF node %d leaves %d subnets\n", i, sub[i] + 1);
}
if (!has)
printf(" No SPF nodes\n");
}

return 0;
}

[/code]