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傳送門

Sol

就相當于從\(x_0\)開始左右橫走，顯然可以設\(f[0/1][i][j]\)表示左到\(i\)右到\(j\)，當前在左/右的代價
但是不好記轉移代價，因為不知道時間
那么可以把它變成最小損失，每次轉移就是加上兩邊沒走的蛋降落的損失，最后用總代價相減即可

# 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; }

轉載于:https://www.cnblogs.com/cjoieryl/p/8319819.html

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