38 lines
1.3 KiB
C++
38 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 55;
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int n, m;
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int w[N][N];
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int f[N * 2][N][N];
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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cin >> w[i][j];
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// 左上角是(1,1),k表示两个小朋友所在位置的x+y的和，最多是n+m
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for (int k = 2; k <= n + m; k++)
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for (int x1 = 1; x1 <= n; x1++) // 第一个小朋友竖着走的距离
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for (int x2 = 1; x2 <= n; x2++) { // 第二个小朋友竖着走的距离
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int y1 = k - x1, y2 = k - x2; // 计算横着走的距离
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// 不能出界，只走有效的位置
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if (y1 < 1 || y1 > m || y2 < 1 || y2 > m) continue;
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// 将本位置的数值加上
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int &x = f[k][x1][x2];
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x = max(x, f[k - 1][x1 - 1][x2] + w[x1][y1]);
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x = max(x, f[k - 1][x1 - 1][x2 - 1] + w[x1][y1]);
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x = max(x, f[k - 1][x1][x2 - 1] + w[x1][y1]);
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x = max(x, f[k - 1][x1][x2] + w[x1][y1]);
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// 如果不是重复的位置，还可以继续加上
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if (x1 != x2) f[k][x1][x2] += w[x2][y2];
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}
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// 输出DP的结果
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printf("%d\n", f[n + m][n][n]);
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return 0;
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}
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