BZOJ3622 已经没有什么好害怕的了【dp + 二项式反演】-优快云博客

本文针对BZOJ3622问题提供两种解题思路：一是利用二项式反演进行求解；二是采用直观的方法，通过求出部分中间结果再进行计算。文章详细介绍了每种方法的具体实现过程。

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题目链接

BZOJ3622

题解

既已开题
~~那就已经没有什么好害怕的了~~

由题目中奇怪的条件我们可以特判掉\(n - k\)为奇数时答案为\(0\)
否则我们要求的就是糖果大于药片恰好有\(\frac{n - k}{2} + k\)个的方案数，我们记为\(K\)

思路1

直接求恰好不好求，想到二项式反演：
如果有
\[b_k = \sum\limits_{i = k}^{n} {i \choose k} a_i\]
那么有
\[a_k = \sum\limits_{i = k}^{n} (-1)^{i - k} {i \choose k} b_i\]

如果我们令\(a_k\)为恰好有\(K\)对糖果大于药片的方案数
从式子来看，对于每一种\(a_i\)的情况，\(b_k\)都从中选出任意\(k\)个
那么\(b_k\)就是任选\(k\)个满足条件，剩余随便选的方案数

问题就转化为了求出\(b_k\)，显然可以\(dp\)
我们先将糖果与药片分别升序排序
我们设\(f[i][j]\)表示前\(i\)个糖果选出\(j\)个和药片匹配，都大于药片的方案数
我们记\(pos_i\)为第\(i\)个糖果比药片大的最大的位置，那么如果第\(i\)个糖果参与匹配，就有\(pos_i\)种选择
由于糖果是升序排序，所以剩余\(j - 1\)个糖果选择范围一定不比\(i\)大，所以\(i\)实际剩余\(pos_i - (j - 1)\)种选择
那么有
\[f[i][j] = f[i - 1][j] + f[i - 1][j - 1] * (pos_i - j + 1)\]

求出\(f[n][i]\)，则\(b_k = f[n][k] * (n - k)!\)
然后再用二项式反演即可求出\(a_k\)

思路2

如果你不知道二项式反演，其实有一个更直观的方法
同样是求出\(f[n][i]\)后求出\(b_i\)
我们令\(ans[k]\)表示恰好\(k\)个糖果大于药片的方案数
显然
\[ans[n] = b_n\]
当\(k < n\)时，\(b_k\)多算了恰好为\(k + 1\)、\(k + 2\)、\(k + 3......\)的方案数，减去即可
\[ans[k] = b_k - \sum\limits_{i = k + 1}^{n} ans[i]\]
这样也是对的

PS

由思路2其实我们可以看到二项式反演实际上就是一个容斥的结果

思路一

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 1000000000,P = 1000000009;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
LL f[maxn][maxn],C[maxn][maxn],fac[maxn];
int a[maxn],b[maxn];
int n,K;
void init(){
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
    for (int i = 0; i <= n; i++){
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j <= (i >> 1); j++)
            C[i][j] = C[i][i - j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
    }
    sort(a + 1,a + 1 + n);
    sort(b + 1,b + 1 + n);
}
int main(){
    n = read(); K = read();
    if ((n - K) & 1) {puts("0"); return 0;}
    K = ((n - K) >> 1) + K;
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 1; i <= n; i++) b[i] = read();
    init();
    f[0][0] = 1;
    for (int i = 1; i <= n; i++){
        f[i][0] = f[i - 1][0];
        int pos = n;
        while (pos && b[pos] >= a[i]) pos--;
        for (int j = 1; j <= i; j++)
            f[i][j] = (f[i - 1][j] + f[i - 1][j - 1] * max(pos - j + 1,0) % P) % P;
    }
    LL ans = 0,t;
    for (int i = K; i <= n; i++){
        t = ((i - K) & 1) ? -1 : 1;
        ans = ((ans + t * C[i][K] % P * (f[n][i] * fac[n - i] % P) % P) % P + P) % P;
    }
    printf("%lld\n",ans);
    return 0;
}

思路2

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 1000000000,P = 1000000009;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
LL f[maxn][maxn],C[maxn][maxn],fac[maxn],ans[maxn];
int a[maxn],b[maxn];
int n,K;
void init(){
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
    for (int i = 0; i <= n; i++){
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j <= (i >> 1); j++)
            C[i][j] = C[i][i - j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
    }
    sort(a + 1,a + 1 + n);
    sort(b + 1,b + 1 + n);
}
int main(){
    n = read(); K = read();
    if ((n - K) & 1) {puts("0"); return 0;}
    K = ((n - K) >> 1) + K;
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 1; i <= n; i++) b[i] = read();
    init();
    f[0][0] = 1;
    for (int i = 1; i <= n; i++){
        f[i][0] = f[i - 1][0];
        int pos = n;
        while (pos && b[pos] >= a[i]) pos--;
        for (int j = 1; j <= i; j++)
            f[i][j] = (f[i - 1][j] + f[i - 1][j - 1] * max(pos - j + 1,0) % P) % P;
    }
    ans[n] = f[n][n];
    for (int i = n - 1; i >= K; i--){
        ans[i] = f[n][i] * fac[n - i] % P;
        for (int j = i + 1; j <= n; j++)
            ans[i] = (ans[i] - C[j][i] * ans[j] % P) % P;
        ans[i] = (ans[i] + P) % P;
    }
    printf("%lld\n",ans[K]);
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/9028747.html