hdu4336 Card Collector MinMax 容斥

题目传送门

https://vjudge.net/problem/HDU-4336

http://acm.hdu.edu.cn/showproblem.php?pid=4336

题解

minmax 容斥模板题。

一个集合 \(S\) 的至少有一个邮票出现的最早时间是 \(\frac 1{\sum\limits_{i\in S} p_i}\)。

时间复杂度 \(O(2^n)\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

#define lowbit(x) ((x) & -(x))

const int N = 20 + 7;
const int NP = (1 << 20) + 7;

int n, S;
double p[N], f[NP];
int pcnt[NP];

inline void work() {
    S = (1 << n) - 1;
    double ans = 0;
    for (int s = 1; s <= S; ++s) f[s] = f[s ^ lowbit(s)] + p[std::__lg(lowbit(s)) + 1], pcnt[s] = pcnt[s ^ lowbit(s)] + 1;
    for (int s = 1; s <= S; ++s)
        if (pcnt[s] & 1) ans += 1 / f[s];
        else ans -= 1 / f[s];
    printf("%.10lf\n", ans);
}

inline void init() {
    for (int i = 1; i <= n; ++i) scanf("%lf", &p[i]);
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    while (~scanf("%d", &n) && n) {
        init();
        work();
    }
    fclose(stdin), fclose(stdout);
    return 0;
}

标签：__,hdu4336,容斥,typedef,long,template,MinMax,inline,define
来源： https://www.cnblogs.com/hankeke/p/hdu4336.html