牛客练习赛6 C 手铐

手铐

思路：

缩环+树形dp

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 emplace_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 = 1e6 + 5;
const int MOD = 19260817;
vector<int> g[N], G[N];
int bcc[N], sz[N], Cnt[N];
int dp[N];
bool vis[N];
int low[N], dfn[N], stk[N], tot = 0, top = 0, cnt = 0;
inline void tarjan(int u, int fa) {
    low[u] = dfn[u] = ++tot;
    stk[++top] = u;
    vis[u] = true;
    for (int v : g[u]) {
        if(v == fa) continue;
        if(!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u]) {
        ++cnt;
        while(stk[top] != u) vis[stk[top]] = false, bcc[stk[top--]] = cnt;
        vis[stk[top]] = false, bcc[stk[top--]] = cnt;
    }
}
int n, m, u, v;
int ans;
inline void dfs(int u, int o) {
    if(Cnt[u] != 1) dp[u] = 1;
    else dp[u] = 0;
    for (int v : G[u]) {
        if(v == o) continue;
        dfs(v, u);
        ans = (ans + dp[u]*1LL*dp[v]%MOD) % MOD;
        if(Cnt[u] != 1) dp[u] = (dp[u] + dp[v]*2) % MOD;
        else dp[u] = (dp[u] + dp[v]) % MOD;
    }
}
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, 0);
    for (int i = 1; i <= n; ++i) {
        for (int v : g[i]) {
            if(bcc[i] == bcc[v]) continue;
            G[bcc[i]].pb(bcc[v]);
        }
        Cnt[bcc[i]]++;
    }
    dfs(1, 0);
    ans = (ans + MOD) % MOD;
    printf("%d\n", ans);
    return 0;
}