题面

传送门

Sol

就相当于从\(x_0\)开始左右横走，显然可以设\(f[0/1][i][j]\)表示左到\(i\)右到\(j\)，当前在左/右的代价
但是不好记转移代价，因为不知道时间
那么可以把它变成最小损失，每次转移就是加上两边没走的蛋降落的损失，最后用总代价相减即可

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1010);
const ll INF(1e18);

IL ll Read(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, x0;
struct Egg{
    ll x, y, v;
    IL bool operator <(RG Egg B) const{  return x < B.x;  }
} w[_];
ll f[2][_][_], sum[_], s;

int main(RG int argc, RG char* argv[]){
    n = Read(); x0 = Read();
    for(RG int i = 1; i <= n; ++i) w[i].x = Read();
    for(RG int i = 1; i <= n; ++i) w[i].y = Read();
    for(RG int i = 1; i <= n; ++i) w[i].v = Read();
    w[++n] = (Egg){x0, 0, -INF}; sort(w + 1, w + n + 1);
    for(RG int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + (w[i].v == -INF ? 0 : w[i].v), s += w[i].y;
    for(x0 = 1; x0 <= n; ++x0) if(w[x0].v == -INF) break;
    Fill(f, 63); f[0][x0][x0] = f[1][x0][x0] = 0;
    for(RG int len = 1; len < n; ++len)
        for(RG int l = max(1, x0 - len); l <= x0; ++l){
            RG int r = l + len; if(r > n) break;
            f[0][l][r] = min(f[0][l][r], f[0][l + 1][r] + (sum[l] + sum[n] - sum[r]) * (w[l + 1].x - w[l].x));
            f[0][l][r] = min(f[0][l][r], f[1][l + 1][r] + (sum[l] + sum[n] - sum[r]) * (w[r].x - w[l].x));
            f[1][l][r] = min(f[1][l][r], f[0][l][r - 1] + (sum[l - 1] + sum[n] - sum[r - 1]) * (w[r].x - w[l].x));
            f[1][l][r] = min(f[1][l][r], f[1][l][r - 1] + (sum[l - 1] + sum[n] - sum[r - 1]) * (w[r].x - w[r - 1].x));
        }
    printf("%.3lf\n", 1.0 * (s - min(f[0][1][n], f[1][1][n])) / 1000);
    return 0;
}