CF248E Piglet's Birthday(概率dp)

题面

luogu
CodeForces

题解

\(orz\) yyb

转移蜜罐其实是吓唬人的......
转移的蜜罐都是尝试过的,所有只关心当前架子上的蜜罐数

\(f[i][j]\)表示第i个货架有j个蜜罐没有被试吃的概率
然后枚举品尝了几个之前没有品尝过的,组合数转移一下

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 100010;
int a[N], b[N];
double f[N][110];

inline LL C(int n, int m) {
    if (n < m) return 0;
    LL s = 1;
    for (int i = n; i > n-m; i--) s *= i;
    for (int i = m; i > 1; i--) s /= i;
    return s;
}

int main() {
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    int n, q; read(n);
    for (int i = 1; i <= n; i++) read(a[i]), b[i] = a[i];
    double ans = 0;
    for (int i = 1; i <= n; i++) f[i][a[i]] = 1;
    for (int i = 1; i <= n; i++) ans += f[i][0];
    read(q);
    while (q--) {
        int u, v, K; read(u); read(v); read(K);
        ans -= f[u][0];
        for (int i = 0; i <= a[u]; i++) {
            double g = 0;
            LL t = C(b[u], K);
            for (int k = 0; k <= K; k++) g += f[u][i+k]*C(i+k, k)*C(b[u]-i-k, K-k)*1.0/t;
            f[u][i] = g;
        }
        b[u] -= K; b[v] += K;
        ans += f[u][0];
        printf("%.10lf\n", ans);
    }
    return 0;
}

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转载自www.cnblogs.com/zzy2005/p/10220190.html