4316: 小C的独立集
思路：先将树上的转移做好。然后环上的转移就是强制最上面的的点选或者不选，然后在环上跑一遍转移就可以了。
代码：
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
const int N = 6e4 + 10;
vector<int> g[N];
int n, m, u, v, dp[N][2], fa[N];
int low[N], dfn[N], stk[N], cmp[N], cnt = 0, top = 0, tot = 0;
bool vis[N];
inline void DP(int u, int v) {
int now0 = 0, now1 = 0;
for (int i = v; i != u; i = fa[i]) {
int t0 = now0 + dp[i][0];
int t1 = now1 + dp[i][1];
now0 = max(t0, t1);
now1 = t0;
}
dp[u][0] += now0;
now0 = 0, now1 = -10000000;
for (int i = v; i != u; i = fa[i]) {
int t0 = now0 + dp[i][0];
int t1 = now1 + dp[i][1];
now0 = max(t0, t1);
now1 = t0;
}
dp[u][1] += now1;
}
inline void tarjan(int u) {
low[u] = dfn[u] = ++cnt;
stk[++top] = u;
vis[u] = true; //标记是否在栈中
dp[u][1] = 1;
for (int i = 0; i < g[u].size(); ++i) {
int v = g[u][i];
if(!dfn[v]) {
fa[v] = u;
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(vis[v]) low[u] = min(low[u], dfn[v]);
if(low[v] > dfn[u]) {
dp[u][0] += max(dp[v][0], dp[v][1]);
dp[u][1] += dp[v][0];
}
}
for (int i = 0; i < g[u].size(); ++i) {
int v = g[u][i];
if(fa[v] != u && dfn[v] > dfn[u]) {
DP(u, v);
}
}
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= m; ++i) scanf("%d %d", &u, &v), g[u].pb(v), g[v].pb(u);
tarjan(1);
printf("%d\n", max(dp[1][0], dp[1][1]));
return 0;
}