PKUWC 2018 Day 2 简要题解

*注意：题面请移步至loj查看。

从这里开始

• Problem A 随机算法
• Problem B 猎人杀
• Problem C 随机游走

　　怎么PKU和THU都编了一些假算法，然后求正确率[汗]。

　　之前听说PKUWC 2017考了五道dp，看来是真的。[完蛋.jpg]

　　这三道题，2道概率，1道期望，和佬题组成一样。。。

　　（我只是把PKUWC 2017在loj后3题做了，是不是全是Day 2就不知道了）

Problem A 随机算法

题目大意

　　给定一个图。随机一个排列$p_{1}p_2p_3cdots p_n$。假设当前有一个独立集集合$S$，从小到大枚举$i$，尝试将$p_i$加入$S$中。问得到最大独立集的概率。

　　每个点有三个状态：被考虑其在独立集中，被考虑且不再独立集中，还未被考虑。

　　设$f_S$表示，当独立集状态为$S$的概率。

　　每次我们选择独立集中的点的时候，把新增加的不可能在独立集中的点在排列中的位置也考虑了。

　　这样一个状态，我们就能表示上面三种点了。

　　时间复杂度$O(n2^n)$

Code

/**
 * loj
 * Problem#2540
 * Accepted
 * Time: 623ms
 * Memory: 10876k
 */
#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
typedef bool boolean;

const int Mod = 998244353;

const int N = 20, S = 1 << N;

int add(int a, int b) {
    return ((a += b) >= Mod) ? (a - Mod) : (a);
}

int mul(int a, int b) {
    return a * 1ll * b % Mod;
}

void exgcd(int a, int b, int& x, int& y) {
    if (!b)
        x = 1, y = 0;
    else {
        exgcd(b, a % b, y, x);
        y -= (a / b) * x; 
    }
}

int inv(int a, int n) {
    int x, y;
    exgcd(a, n, x, y);
    return (x < 0) ? (x + n) : (x);
}

int n, m;
int bit[S];
int g[N], f[S];
boolean indep[S];
int fac[N], _fac[N], adj[S];

inline void init() {
    scanf("%d%d", &n, &m);
    for (int i = 0, u, v; i < m; i++) {
        scanf("%d%d", &u, &v);
        --u, --v;
        g[u] |= (1 << v);
        g[v] |= (1 << u);
    }
}

int permu(int n, int m) {
    if (n < m)
        return 0;
    return mul(fac[n], _fac[n - m]);
}

inline void solve() {
    int all = 1 << n;
    indep[0] = true;
    for (int i = 1; i < all; i++)
        for (int j = 0; j < n && !indep[i]; j++)
            if (((i >> j) & 1) && indep[i ^ (1 << j)] && !(g[j] & i))
                indep[i] = true;
    
    for (int i = 1; i < all; i++)
        bit[i] = bit[i >> 1] + (i & 1);
    
    int mx = 0;
    for (int i = 1; i < all; i++)
        if (indep[i] && bit[i] > mx)
            mx = bit[i];
        
    adj[0] = 0;
    for (int i = 1; i < all; i++)
        for (int j = 0; j < n && !adj[i]; j++)
            if ((i >> j) & 1)
                adj[i] = adj[i ^ (1 << j)] | g[j];    

    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = mul(fac[i - 1], i);
    _fac[n] = inv(fac[n], Mod);
    for (int i = n; ~i; i--)
        _fac[i - 1] = mul(_fac[i], i);
    
    f[0] = 1;
    for (int s = 0; s < all - 1; s++) {
        if (!f[s] || !indep[s])
            continue;
        int ban = adj[s] | s;
        for (int i = 0; i < n; i++) {
            if (((ban >> i) & 1))
                continue;
            f[s | (1 << i)] = add(f[s | (1 << i)], mul(f[s], permu(n - bit[ban] - 1, bit[g[i]] - bit[ban & g[i]])));
        }
    }
    int res = 0;
    for (int i = 1; i < all; i++)
        if (indep[i] && bit[i] == mx)
            res = add(res, f[i]);
    res = mul(res, _fac[n]);
    printf("%d
", res);
}

int main() {
    init();
    solve();
    return 0;
}

Problem B 猎人杀

题目大意

　　（原题题意过于简洁不需要大意）

　　考虑求出在$1$被消灭掉后，剩下的集合为$S$的概率$f_S$。

　　发现直接求它不好求，考虑求出剩下的集合包含$S$时的概率$g_S$。

　　我们希望求$f_{varnothing}$。

　　我们发现这个分母的变化会使得事情变得毒瘤起来，考虑被打死的人不会退出游戏，这样不会改变要求的概率。

　　设$A = sum_{i = 1}^{n}w_i$，$B = sum_{i in S}w_i$，那么有：

$egin{align} g_s &= frac{w_1}{A}sum_{i = 0}^{infty}left ( 1 - frac{B + w_1}{A} ight )^{i} \ &= frac{w_{1}}{A}cdot frac{1}{1 - left(1 - frac{B +w_1}{A} ight )} \ &= frac{w_1}{B + w_1 }end{align}$

　　因为剩下的集合不同的两个事件是互斥的，所以有：

$egin{align} g_s = sum_{tsubseteq Uwedge ssubseteq t}f_{t}end{align}$

　　反演得到：

$egin{align} f_s = sum_{tsubseteq Uwedge ssubseteq t} (-1)^{|t| - |s|}g_tend{align}$

　　因为$w$的和不超过$10^5$，考虑求出项$frac{w_{1}}{w_{1} + B}$的系数。

　　直接构造一个函数$G(x) = prod_{i = 2}^{n}(1 - x^{w_i})$，用分治NTT求出来就好了。

Code

/**
 * loj
 * Problem#2541
 * Accepted
 * Time: 940ms
 * Memory: 7084k
 */
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;
#define ll long long
using std::reverse;

// Polynomial Template

template <typename T>
void pfill(T* pst, const T* ped, T val) {
    for ( ; pst != ped; *(pst++) = val);
}

template <typename T>
void pcopy(T* pst, const T* ped, T* pval) {
    for ( ; pst != ped; *(pst++) = *(pval++));
}

const int N = 262144;
const int M = 998244353;
const int inv2 = (M + 1) >> 1;
const int g = 3;
const int bzmax = 18;

void debug(int *f, int len) {
    for (int i = 0; i < len; i++)
        std :: cerr << f[i] << " ";
    std :: cerr << '
';
}

int add(int a, int b) {
    return ((a += b) >= M) ? (a - M) : (a);
}

int sub(int a, int b) {
    return ((a -= b) < 0) ? (a + M) : (a);
}

int mul(int a, int b) {
    return a * 1ll * b % M;
}

int qpow(int a, int p) {
    if (p < 0)
        p += M - 1;
    int rt = 1, pa = a;
    for ( ; p; p >>= 1, pa = pa * 1ll * pa % M)
        if (p & 1)
            rt = rt * 1ll * pa % M;
    return rt;
}

class NTT {
    public:
        int gn[1 << (bzmax + 2)], rgn[1 << (bzmax + 2)];
        
        NTT() {
            int *p = gn, *_p = rgn;
            for (int i = 0, l = 2; i < bzmax; i++, l <<= 1) {
//                gn[i] = qpow(g, (M - 1) / l);
//                rgn[i] = qpow(g, -(M - 1) / l);
                int tmp = 1;
                int wn = qpow(g, (M - 1) >> (i + 1));
                int _wn = qpow(g, - ((M - 1) >> (i + 1)));
                for (int j = 0; j < l; j++)
                    *(p++) = tmp, tmp = mul(tmp, wn);
                tmp = 1;
                for (int j = 0; j < l; j++)
                    *(_p++)= tmp, tmp = mul(tmp, _wn);
//                cerr << i << " " << cnt << '
';
            }
        }
        
        void rader(int *f, int len) {
            for (int i = 1, j = len >> 1, k; i < len - 1; i++, j += k) {
                if (i < j)
                    std::swap(f[i], f[j]);
                for (k = len >> 1; j >= k; j -= k, k >>= 1);
            }
        }

        void operator () (int* f, int len, int sign) {
            rader(f, len);
            const int *p = ((sign > 0) ? (gn) : (rgn));
            for (int b = 2; b <= len; b <<= 1) {
//                int wn = ((sign > 0) ? (gn[t]) : (rgn[t])), w = 1, hb = b >> 1;
                const int * const pc = p;
                const int hb = b >> 1;
                for (int i = 0; i < len; i += b, p = pc)
                    for (int j = i; j < i + hb; j++) {
                        int a = f[j], b = *p++ * 1ll * f[j + hb] % M;
                        f[j] = add(a, b);
                        f[j + hb] = sub(a, b);
                    }
                p = p + b;
            }

            if (sign == -1) {
                int invlen = qpow(len, M - 2);
                for (int i = 0; i < len; i++)
                    f[i] = f[i] * 1ll * invlen % M;
            }
        }

        int correctLen(int n) {
            int t = 1;
            for ( ; t < n; t <<= 1);
            return t;
        }

}NTT;

typedef vector<int> Poly;

Poly operator * (Poly a, Poly b) {
    int n = (signed) (a.size() + b.size()) - 1;
    if (n < 64) {
        Poly rt (n);
        pfill(rt.data(), rt.data() + n, 0);
        for (unsigned i = 0; i < a.size(); i++)
            for (unsigned j = 0; j < b.size(); j++)
                rt[i + j] = add(rt[i + j], mul(a[i], b[j]));
        return rt;
    }
    int t = NTT.correctLen(n);
    Poly rt (t);
    a.resize(t), b.resize(t);
    NTT(a.data(), t, 1);
    NTT(b.data(), t, 1);
    for (int i = 0; i < t; i++)
        rt[i] = mul(a[i], b[i]);
    NTT(rt.data(), t, -1);
    rt.resize(n);
    return rt;
}

Poly dividing(int n, int* a) {
    if (n == 1) {
        Poly rt (*a + 1);
        pfill(rt.data(), rt.data() + *a + 1, 0);
        rt[*a] = M - 1, rt[0] = 1;
        return rt;
    }
    int sum = 0, L = 0, s = *a;
    for (int i = 0; i < n; i++)
        sum += a[i];
    sum >>= 1;
    for (int i = 1; i < n; s += a[i], i++)
        if (s + a[i] > sum) {
            L = i;
            break;
        }
    return dividing(L, a) * dividing(n - L, a + L);
}

int n, s;
int *w;

inline void init() {
    scanf("%d", &n);
    w = new int[(n + 1)];
    for (int i = 0; i < n; i++) {
        scanf("%d", w + i);
        if (i > 0)
            s += w[i];
    }
}

inline void solve() {
    if (n == 1) {
        puts("1");
        return;
    }
    sort(w + 1, w + n);
    Poly p = dividing(n - 1, w + 1);
    int res = 1;
    for (int i = 1; i <= s; i++)
        if (p[i])
            res = add(res, mul(mul(w[0], p[i]), qpow(add(w[0], i), -1)));    
    printf("%d
", res);
}

int main() {
    init();
    solve();
    return 0;
}

Problem C 随机游走

题目大意

　　（原题题意过于简洁不需要大意）

Solution 1

　　设$f_{s, i}$表示当前在点$i$，走完$s$中包含的点期望还需要多少步。

　　对于非根节点的方程可以化为$f_{s,i} = kcdot f_{s', fa_i} + b$的形式，然后两遍dfs就行了。

Code

/**
 * loj
 * Problem#2542
 * Accepted
 * Time: 997ms
 * Memory: 37116k
 */
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <vector>
using namespace std;
typedef bool boolean;

const int Mod = 998244353;

int add(int a, int b) {
    return ((a += b) >= Mod) ? (a - Mod) : (a); 
}

int sub(int a, int b) {
    return ((a -= b) < 0) ? (a + Mod) : (a);
}

int mul(int a, int b) {
    return a * 1ll * b % Mod;
}

void exgcd(int a, int b, int& x, int& y) {
    if (!b)
        x = 1, y = 0;
    else {
        exgcd(b, a % b, y, x);
        y -= (a / b) * x;
    }
}

int inv(int a, int n) {
    int x, y;
    exgcd(a, n, x, y);
    return (x < 0) ? (x + n) : (x);
}

const int N = 18, S = 1 << N;

typedef class Value {
    public:
        int k, b;

        Value(int k = 0, int b = 0) : k(k), b(b) {    }
}Value;

int n, q, X;
int msk_all;
Value f[S][N];
vector<int> g[N];

inline void init() {
    scanf("%d%d%d", &n, &q, &X), --X;
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        --u, --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    msk_all = (1 << n) - 1;
}

void dfs1(int p, int S, int Fa) {
    int deg = (signed) g[p].size(), coef = deg, cont = deg, nS;
    for (int i = 0, e; i < deg; i++) {
        if ((e = g[p][i]) == Fa)
            continue;
        nS = S & (msk_all ^ (1 << e));
        dfs1(e, S, p);
        if (nS ^ S) {
            cont = add(cont, f[nS][e].b);
        } else {
            coef = sub(coef, f[S][e].k);
            cont = add(cont, f[S][e].b);
        }
    }
        
    if ((1 << p) & S)
        return;

    int _coef = inv(coef, Mod);
    if (p == X) {
        f[S][p] = Value(0, mul(cont, _coef));    
    } else {
        f[S][p] = Value(_coef, mul(_coef, cont));
    }
}    

void dfs2(int p, int S, int Fa) {
    if (p != X && !((1 << p) & S)) {
        int nS = S & (msk_all ^ (1 << Fa));
        f[S][p].b = add(mul(f[S][p].k, f[nS][Fa].b), f[S][p].b);
        f[S][p].k = 0;
    }

    for (int i = 0, e; i < (signed) g[p].size(); i++) {
        if ((e = g[p][i]) == Fa)
            continue;
        dfs2(e, S, p);
    }
}

inline void solve() {
    for (int s = 1; s <= msk_all; s++) {
        dfs1(X, s, -1);
        dfs2(X, s, -1);
    }

    int K, msk;    
    while (q--) {
        scanf("%d", &K);
        msk = 0;
        for (int i = 0, x; i < K; i++)
            scanf("%d", &x), msk |= (1 << (x - 1));
        msk = msk & (msk_all ^ (1 << X));
        printf("%d
", f[msk][X].b);
    }
}

int main() {
    init();
    solve();
    return 0;
}

Solution 2

　　用最值反演把问题转化成求走到$s$中任意一个点的期望。

　　假设当前考虑的集合是$s$，设$f_i$表示当前在点$i$，走到$s$中任意一个点的期望。

　　然后同样的做法dp一下就行了。

Code

/**
 * loj
 * Problem#2542
 * Accepted
 * Time: 617ms
 * Memory: 1232k
 */
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <vector>
using namespace std;
typedef bool boolean;

const int Mod = 998244353;

int add(int a, int b) {
    return ((a += b) >= Mod) ? (a - Mod) : (a); 
}

int sub(int a, int b) {
    return ((a -= b) < 0) ? (a + Mod) : (a);
}

int mul(int a, int b) {
    return a * 1ll * b % Mod;
}

void exgcd(int a, int b, int& x, int& y) {
    if (!b)
        x = 1, y = 0;
    else {
        exgcd(b, a % b, y, x);
        y -= (a / b) * x;
    }
}

int inv(int a, int n) {
    int x, y;
    exgcd(a, n, x, y);
    return (x < 0) ? (x + n) : (x);
}

const int N = 18, S = 1 << N;

typedef class Value {
    public:
        int k, b;

        Value(int k = 0, int b = 0) : k(k), b(b) {    }
}Value;

int n, q, X;
int msk_all;
Value f[N];
int F[S];
vector<int> g[N];

inline void init() {
    scanf("%d%d%d", &n, &q, &X), --X;
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        --u, --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    msk_all = (1 << n) - 1;
}

void dfs1(int p, int S, int Fa) {
    int deg = (signed) g[p].size(), coef = deg, cont = deg, nS;
    for (int i = 0, e; i < deg; i++) {
        if ((e = g[p][i]) == Fa)
            continue;
        nS = S & (1 << e);
        dfs1(e, S, p);
        if (!nS) {
            coef = sub(coef, f[e].k);
            cont = add(cont, f[e].b);
        }
    }
        
    int _coef = inv(coef, Mod);
    if ((1 << p) & S)
        f[p] = Value(0, 0);
    else if (p == X) {
        f[p] = Value(0, mul(cont, _coef));    
    } else {
        f[p] = Value(_coef, mul(_coef, cont));
    }
}    

void dfs2(int p, int S, int Fa) {
    if (p != X && !((1 << p) & S)) {
        f[p].b = add(mul(f[p].k, f[Fa].b), f[p].b);
        f[p].k = 0;
    }

    for (int i = 0, e; i < (signed) g[p].size(); i++) {
        if ((e = g[p][i]) == Fa)
            continue;
        dfs2(e, S, p);
    }
}

inline void solve() {
    for (int s = 1, bit = 0; s <= msk_all; s++) {
        if (s & (1 << X)) {
            F[s] = 0;
            continue;
        }
        dfs1(X, s, -1);
        dfs2(X, s, -1);
        F[s] = f[X].b, bit = 0;
        for (int s0 = s; s0; s0 -= (s0 & (-s0)), bit ^= 1);
        if (!bit)
            F[s] = Mod - F[s];
    }

    for (int i = 0; i < n; i++)
        for (int j = 0; j <= msk_all; j++)
            if ((j >> i) & 1)
                F[j] = add(F[j], F[j ^ (1 << i)]);

    int K, msk;    
    while (q--) {
        scanf("%d", &K);
        msk = 0;
        for (int i = 0, x; i < K; i++)
            scanf("%d", &x), msk |= (1 << (x - 1));
        msk = msk & (msk_all ^ (1 << X));
        printf("%d
", F[msk]);
    }
}

int main() {
    init();
    solve();
    return 0;
}