【BZOJ4455】【UOJ185】【ZJOI2016】小星星（树形DP，容斥原理）

Description

http://uoj.ac/problem/185

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Solution

单考虑下面Code中的那个dp函数，设$f_{i,j}$表示$i$映射到$j$的$i$的子树中的答案，直接$O(n^3)$转移即可。
但是这样是错的，因为有可能很多个点映射到同一个点。
所以可以容斥一下，枚举最多映射哪些点，乘上容斥系数就行了。这题的容斥系数就是常见的$(-1)^k$。

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Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ4455][UOJ185][ZJOI2016] 小星星
 * Date: Jun 6th, 2018
 * Email: [email protected]
 * Institute: Yali High School
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#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 SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

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() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 19;

int n, m, sta, FULL, tbeg[maxn], tnex[maxn * maxn], tv[maxn * maxn], gbeg[maxn], gnex[maxn * maxn], gv[maxn * maxn], te, ge;
LL f[maxn][maxn];

inline void tadd(int uu, int vv) { tv[++ te] = vv, tnex[te] = tbeg[uu], tbeg[uu] = te; }
inline void gadd(int uu, int vv) { gv[++ ge] = vv, gnex[ge] = gbeg[uu], gbeg[uu] = ge; }

void dp(int u, int pa)
{
    for (register int i = tbeg[u]; i; i = tnex[i]) if (tv[i] != pa) dp(tv[i], u);
    For(i, 1, n) if (sta & (1 << (i - 1))) { //Enumerate the node matched u
        f[u][i] = 1;
        for (register int j = tbeg[u]; j; j = tnex[j]) if (tv[j] != pa) { //Enumerate sons of u
            LL sum = 0;
            for (register int k = gbeg[i]; k; k = gnex[k]) sum += f[tv[j]][gv[k]]; //Enumerate the node matched sons of u
            f[u][i] *= sum;
        }
    }
}

int main()
{
#ifdef hany01
    File("bzoj4455");
#endif

    static int uu, vv;
    static LL ans = 0;

    n = read(), m = read();
    For(i, 1, m) uu = read(), vv = read(), gadd(uu, vv), gadd(vv, uu);
    For(i, 2, n) uu = read(), vv = read(), tadd(uu, vv), tadd(vv, uu);

    for (sta = 1, FULL = 1 << n; sta < FULL; ++ sta) {
        Set(f, 0), dp(1, 0);
        LL sum = 0;
        For(i, 1, n) sum += f[1][i];
        ans += sum * (((n - __builtin_popcount(sta)) & 1) ? -1 : 1);
    }
    printf("%lld\n", ans);

    return 0;
}