CF912D Fishes 期望 + 贪心

有趣的水题

由期望的线性性质，全局期望 = 每个格子的期望之和

由于权值一样，我们优先选概率大的点就好了

用一些数据结构来维护就好了

复杂度$O(k \log n)$

#include <set>
#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    #define gc getchar
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();
        return p * w;
    }
    int wr[], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('');
        if(o < ) o = -o, pc('-');
        while(o) wr[++ rw] = o % , o /= ;
        while(rw) pc(wr[rw --] + '');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b,  : ; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b,  : ; }
}
using namespace std;
using namespace remoon;

namespace mod_mod {
    #define mod 1000000007
    inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
    inline void dec(int &a, int b) { a -= b; if(a < ) a += mod; }
    inline int Inc(int a, int b) { return (a + b >= mod) ? a + b - mod : a + b; }
    inline int Dec(int a, int b) { return (a - b < ) ? a - b + mod : a - b; }
    inline int mul(int a, int b) { return 1ll * a * b % mod; }
}
using namespace mod_mod;

de ans = ;
int n, m, r, k;
int nx[] = { , -, ,  };
int ny[] = { , , , - };
struct node {
    int x, y; ll c;
    friend bool operator < (node a, node b)
    { return a.c < b.c; }
};
priority_queue <node> q;
map <pii, int> vis;

inline bool ck(int x, int y) {
    if(x <  || x > n) return ;
    if(y <  || y > m) return ;
    if(vis[mp(x, y)]) return ;
    else return ;
}

inline ll solve(int x, int y) {
    ll X = min(min(x, n - x + ), min(n - r + , r));
    ll Y = min(min(y, m - y + ), min(m - r + , r));
    return X * Y;
}

int main() {

    n = read(); m = read();
    r = read(); k = read();

    vis[mp(r, r)] = ;
    q.push((node){r, r, solve(r, r)});

    while(k --) {
        node now = q.top(); q.pop();
        int x = now.x, y = now.y; ans += now.c / (de)(n - r + ) / (de)(m - r + );
        rep(i, , ) {
            int dx = x + nx[i], dy = y + ny[i];
            if(ck(dx, dy)) {
                vis[mp(dx, dy)] = ;
                q.push((node){dx, dy, solve(dx, dy)});
            }
        }
    }

    printf("%.9lf\n", ans);
    return ;
}