【BZOJ4316】小C的独立集（仙人掌，DP）

Description

给定一个仙人掌，求最大独立集。

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Solution

对于圆方树上的圆点和圆点之间的连边，我们和树形DP类似的转移即可。
对于每一个方点，我们将其代表的环拿出来处理，在环上跑一个简单的DP、将信息都转移到环的顶部即可。

这个题目不需要把圆方树建出来，只需要在原来建圆方树时找到环后与方点连边的部分改成DP即可。

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Code

/************************************************
 * Au: Hany01
 * Date: Jul 29th, 2018
 * Prob: BZOJ4316 小C的独立集
 * Email: [email protected]
 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 5e4 + 5, maxm = (6e4 + 5) * 2;

int n, beg[maxn], nex[maxm], v[maxm], e = 1, f[2][maxn], fa[maxn], dfn[maxn], low[maxn], clk;

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

inline void Solve(int u, int anc)
{
    static int f0, f1, tmp; f0 = f1 = 0;
    for (int t = u; t != anc; t = fa[t])
        tmp = f1, f1 = f[0][t] + f0, f0 = max(f1, f[1][t] + tmp);
    f[0][anc] += f0, f0 = 0, f1 = -INF;
    for (int t = u; t != anc; t = fa[t])
        tmp = f1, f1 = f[0][t] + f0, f0 = max(f1, f[1][t] + tmp);
    f[1][anc] += f1;
}

void DFS(int u, int pa = 0)
{
    dfn[u] = low[u] = ++ clk, fa[u] = pa, f[1][u] = 1;
    for (register int i = beg[u]; i; i = nex[i]) {
        if (!dfn[v[i]]) DFS(v[i], u), chkmin(low[u], low[v[i]]);
        else if (pa != v[i]) chkmin(low[u], dfn[v[i]]);
        if (dfn[u] < low[v[i]]) f[1][u] += f[0][v[i]], f[0][u] += max(f[0][v[i]], f[1][v[i]]);
    }
    for (register int i = beg[u]; i; i = nex[i])
        if (fa[v[i]] != u && dfn[u] < dfn[v[i]]) Solve(v[i], u);
}

int main()
{
#ifdef hany01
    freopen("bzoj4316.in", "r", stdin);
    freopen("bzoj4316.out", "w", stdout);
#endif

    static int uu, vv, m;

    n = read(), m = read();
    For(i, 1, m) uu = read(), vv = read(), add(uu, vv), add(vv, uu);
    DFS(1), printf("%d\n", max(f[0][1], f[1][1]));

    return 0;
}