ABC 210 D - National Railway

D - National Railway

矩阵DP，拆绝对值

设向下为 $x$ 正方向，向右为 $y$ 正方向

设两个点为 $(x_1,y_1),\;(x_2,y_2)$, 代价为 $A_{x_1y_1}+A_{x_2y_2}+|x_1-x_2|+|y_1-y_2|$

想办法拆掉绝对值

1. $(x_2,y_2)$ 在 $(x_1,y_1)$ 右下方，则 $|x_1-x_2|+|y_1-y_2|=(x_2+y_2)-(x_1+y_1)$

   $cost=A_{x_2y_2}+(x_2+y_2)+(A_{x_1y_1}-(x_1+y_1))$

2. $(x_2,y_2)$ 在 $(x_1,y_1)$ 左下方，则 $|x_1-x_2|+|y_1-y_2|=(x_2-y_2)-(x_1-y_1)$

   $cost=A_{x_2y_2}+(x_2-y_2)+(A_{x_1y_1}-(x_1-y_1))$

在第一种方案中，设 $f[i][j]$ 以 $(i,j)$ 为右下角的小矩阵中，$A_{xy}-(x+y)$ 的最小值

$f[i][j]=min({f[i-1][j],f[i][j-1],A_{ij}-(i+j)})$

所以枚举 $(i,j)$ 作为 $(x_2,y_2)$, $(x_1,y_1)$ 的最小代价就是 $f[i][j]$（注意先更新 $f[i][j]=min(f[i-1][j],f[i][j-1])$ 后求出当前的答案，再更新 $f[i][j]=min(f[i][j],A_{ij}-(i+j))$, 因为 $(x_1,y_1)\neq(x_2,y_2)$）

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>

using namespace std;
#define endl "\n"

typedef long long ll;
typedef pair<int, int> PII;

const int N = 1e3 + 10;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int n, m;
ll c, a[N][N];
ll f[N][N];

ll solve1()
{
	ll ans = INF;
	memset(f, 0x3f, sizeof f);
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			f[i][j] = min(f[i-1][j], f[i][j-1]);
			ans = min(ans, a[i][j] + c * (i + j) + f[i][j]);
			f[i][j] = min(f[i][j], a[i][j] - c * (i + j));
		}
	}
	return ans;
}

ll solve2()
{
	ll ans = INF;
	memset(f, 0x3f, sizeof f);
	for (int i = 1; i <= n; i++)
	{
		for (int j = m; j >= 1; j--)
		{
			f[i][j] = min(f[i-1][j], f[i][j+1]);
			ans = min(ans, a[i][j] + c * (i - j) + f[i][j]);
			f[i][j] = min(f[i][j], a[i][j] - c * (i - j));
		}
	}
	return ans;
}
int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> m >> c;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			cin >> a[i][j];
		
	ll ans = min(solve1(), solve2());
	cout << ans << endl;
    return 0;
}