F - Earn to Advance

F - Earn to Advance

Problem Statement

There is a grid with $N$ rows and $N$ columns. Let $(i,j)$ denote the square at the $i$-th row from the top and $j$-th column from the left.

Takahashi is initially at square $(1,1)$ with zero money.

When Takahashi is at square $(i,j)$, he can perform one of the following in one action:

• Stay at the same square and increase his money by $P_{i,j}$.
• Pay $R_{i,j}$ from his money and move to square $(i,j+1)$.
• Pay $D_{i,j}$ from his money and move to square $(i+1,j)$.

He cannot make a move that would make his money negative or take him outside the grid.

If Takahashi acts optimally, how many actions does he need to reach square $(N,N)$?

Constraints

• $2 \leq N \leq 80$
• $1 \leq P_{i,j} \leq 10^9$
• $1 \leq R_{i,j},D_{i,j} \leq 10^9$
• All input values are integers.

---

Input

The input is given from Standard Input in the following format:

$N$
$P_{1,1}$ $\ldots$ $P_{1,N}$
$\vdots$
$P_{N,1}$ $\ldots$ $P_{N,N}$
$R_{1,1}$ $\ldots$ $R_{1,N-1}$
$\vdots$
$R_{N,1}$ $\ldots$ $R_{N,N-1}$
$D_{1,1}$ $\ldots$ $D_{1,N}$
$\vdots$
$D_{N-1,1}$ $\ldots$ $D_{N-1,N}$

Output

Print the answer.

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Sample Input 1

3
1 2 3
3 1 2
2 1 1
1 2
4 3
4 2
1 5 7
5 3 3

Sample Output 1

8

It is possible to reach square $(3,3)$ in eight actions as follows:

• Stay at square $(1,1)$ and increase money by $1$. His money is now $1$.
• Pay $1$ money and move to square $(2,1)$. His money is now $0$.
• Stay at square $(2,1)$ and increase money by $3$. His money is now $3$.
• Stay at square $(2,1)$ and increase money by $3$. His money is now $6$.
• Stay at square $(2,1)$ and increase money by $3$. His money is now $9$.
• Pay $4$ money and move to square $(2,2)$. His money is now $5$.
• Pay $3$ money and move to square $(3,2)$. His money is now $2$.
• Pay $2$ money and move to square $(3,3)$. His money is now $0$.

---

Sample Input 2

3
1 1 1
1 1 1
1 1 1
1000000000 1000000000
1000000000 1000000000
1000000000 1000000000
1000000000 1000000000 1000000000
1000000000 1000000000 1000000000

Sample Output 2

4000000004

 

解题思路

  要求 $(1,1) \to (n,n)$ 的最小操作次数，不妨先考虑最后一次使用第一种操作的位置，假设为 $(i,j)$。为了保证 $(i,j) \to (n,n)$ 不存在金额为负数的情况，需要在 $(i,j)$ 处停留一定的次数。假设 $h(i,j,n,n)$ 表示 $(i,j) \to (n,n)$ 的所有路径中只执行第二、三种操作的最小金额，$g(i,j)$ 表示 $(1,1) \to (i,j)$ 还剩的金额，那么在 $(i,j)$ 处至少要停留 $c = \left\lceil \frac{\max\{{0, \, h(i,j,n,n) - g(i,j)}\}}{p_{i,j}} \right\rceil$ 次。因此当 $(i,j)$ 是最后一次使用第一种操作的位置时，$(i,j) \to (n,n)$ 的最小操作次数就是 $c + n-i + n-j$。

  现在问题就变成了求 $(1,1) \to (i,j)$ 的最小操作次数，显然可以沿用上面的思路。为此可以考虑 dp，定义 $f(i,j)$ 表示 $(1,1) \to (i,j)$ 的最小操作次数，根据最后一次使用第一种操作的位置 $\{ (u,v) \mid 1 \leq u \leq i, 1 \leq v \leq j \}$ 进行状态划分，状态转移方程就是 $f(i,j) = \min\limits_{(u,v)}\left\{{f(u,v) + c + i-u + j-v}\right\}$，其中 $c = \left\lceil \frac{\max\{{0, \, h(u,v,i,j) - g(u,v)}\}}{p_{u,v}} \right\rceil$。

  考虑维护 $g(i,j)$ 的最大值，当 $f(i,j) > f(u,v) + c + i-u + j-v$ 时，直接令 $g(i,j) = g(u,v) + c \cdot p_{u,v} - h(u,v,i,j)$。否则如果 $f(i,j) = f(u,v) + c + i-u + j-v$，$g(i,j) = \max\{{g(i,j), \, g(u,v) + c \cdot p_{u,v} - h(u,v,i,j)}\}$。

  $h(i,j,u,v)$ 可以用 dp 预处理出来，状态转移方程就是 $h(i,j,u,v) = \min\left\{h(i,j,u-1,v) + d_{u-1,v}, \, h(i,j,u,v-1) + r_{u,v-1}\right\}$。

  AC 代码如下，时间复杂度为 $O(n^4)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 85;

int p[N][N], r[N][N], d[N][N];
LL f[N][N], g[N][N], h[N][N][N][N];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &p[i][j]);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j < n; j++) {
            scanf("%d", &r[i][j]);
        }
    }
    for (int i = 1; i < n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &d[i][j]);
        }
    }
    memset(h, 0x3f, sizeof(h));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            h[i][j][i][j] = 0;
            for (int u = i; u <= n; u++) {
                for (int v = j; v <= n; v++) {
                    if (u == i && v == j) continue;
                    h[i][j][u][v] = min(h[i][j][u - 1][v] + d[u - 1][v], h[i][j][u][v - 1] + r[u][v - 1]);
                }
            }
        }
    }
    memset(f, 0x3f, sizeof(f));
    f[1][1] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int u = 1; u <= i; u++) {
                for (int v = 1; v <= j; v++) {
                    LL c = (max(0ll, h[u][v][i][j] - g[u][v]) + p[u][v] - 1) / p[u][v];
                    if (f[i][j] > f[u][v] + c + i - u + j - v) {
                        f[i][j] = f[u][v] + c + i - u + j - v;
                        g[i][j] = g[u][v] + p[u][v] * c - h[u][v][i][j];
                    }
                    else if (f[i][j] == f[u][v] + c + i - u + j - v) {
                        g[i][j] = max(g[i][j], g[u][v] + p[u][v] * c - h[u][v][i][j]);
                    }
                }
            }
        }
    }
    printf("%lld", f[n][n]);
    
    return 0;
}

  上面代码所需的内存是 $400 \text{ MB}$ 左右。还可以把 $h$ 优化到两维。在枚举 $(i,j)$ 的时候定义 $h(u,v)$ 表示 $(u,v) \to (i,j)$ 只执行第二、三种操作的最小金额。状态转移方程就是 $h(u,v) = \min\left\{{h(u+1,v) + d_{u,v}, \, h(u,v+1) + r_{u,v}}\right\}$。

  AC 代码如下，时间复杂度为 $O(n^4)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 85;

int p[N][N], r[N][N], d[N][N];
LL f[N][N], g[N][N], h[N][N];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &p[i][j]);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j < n; j++) {
            scanf("%d", &r[i][j]);
        }
    }
    for (int i = 1; i < n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &d[i][j]);
        }
    }
    memset(f, 0x3f, sizeof(f));
    f[1][1] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            memset(h, 0x3f, sizeof(h));
            h[i][j] = 0;
            for (int u = i; u; u--) {
                for (int v = j; v; v--) {
                    if (u == i && v == j) continue;
                    h[u][v] = min(h[u + 1][v] + d[u][v], h[u][v + 1] + r[u][v]);
                }
            }
            for (int u = 1; u <= i; u++) {
                for (int v = 1; v <= j; v++) {
                    LL c = (max(0ll, h[u][v] - g[u][v]) + p[u][v] - 1) / p[u][v];
                    if (f[i][j] > f[u][v] + c + i - u + j - v) {
                        f[i][j] = f[u][v] + c + i - u + j - v;
                        g[i][j] = g[u][v] + p[u][v] * c - h[u][v];
                    }
                    else if (f[i][j] == f[u][v] + c + i - u + j - v) {
                        g[i][j] = max(g[i][j], g[u][v] + p[u][v] * c - h[u][v]);
                    }
                }
            }
        }
    }
    printf("%lld", f[n][n]);
    
    return 0;
}

 

参考资料

  Editorial - Toyota Programming Contest 2024#3（AtCoder Beginner Contest 344）：https://atcoder.jp/contests/abc344/editorial/9494