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;

}