BZOJ 4316: 小C的独立集

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], cnt = 0;
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, int o) {
    fa[u] = o;
    dp[u][1] = 1;
    dfn[u] = low[u] = ++cnt;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i];
        if(v == o) continue;
        if(!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else 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, 0);
    printf("%d\n", max(dp[1][0], dp[1][1]));
    return 0;
}