Codeforces 229E Gifts 概率dp （看题解)

Gifts

感觉题解写的就是坨不知道什么东西。。

看得这个题解。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m, num1, num2;
int cnt[N];
int need[N];
bool have[N];
vector<PII> vc;

double dp[N][N];

double sum[N];

double calc(int n, int m) {
    return exp(sum[m] + sum[n - m] - sum[n]);
}

int main() {
    for(int i = 1; i < N; i++) sum[i] = sum[i - 1] + log(i);
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        scanf("%d", &cnt[i]);
        for(int j = 1; j <= cnt[i]; j++) {
            int x; scanf("%d", &x);
            vc.push_back(mk(x, i));
        }
    }
    sort(vc.rbegin(), vc.rend());
    int val = vc[n - 1].fi;
    num1 = n; num2 = 0;
    if(SZ(vc) == n || vc[n].fi != val) val = -1;
    if(~val) {
        for(auto& t : vc) {
            if(t.fi > val) need[t.se]++, num1--;
            else if(t.fi == val) have[t.se] = true, num2++;
            else break;
        }
    } else {
        for(int i = 0; i < n; i++) need[vc[i].se]++;
        num1 = 0, num2 = 0;
    }
    dp[0][0] = 1.0;
    for(int i = 1; i <= m; i++) {
        for(int j = 0; j <= i && j <= num1; j++) {
            if(!have[i]) {
                dp[i][j] = dp[i - 1][j] * calc(cnt[i], need[i]);
            }
            else {
                if(j) dp[i][j] = dp[i - 1][j] * calc(cnt[i], need[i]) + dp[i - 1][j - 1] * calc(cnt[i], need[i] + 1);
                else dp[i][j] = dp[i - 1][j] * calc(cnt[i], need[i]);
            }
        }
    }
    printf("%.15f
", dp[m][num1] * calc(num2, num1));
    return 0;
}

/*
*/