[HDU 2686] Matrix

On the Internet, the question is a charge flow.

I gave a rough look at the data Scope and thought that the topic seemed to make us look like "Big (d) Fart (p)". To respect the speaker, I 'd like to thank you ~

 

First, a loop can be seen as two steps in parallel. This is the common idea of big fart and Fei yongliu.

Therefore, we can use several diagonal lines (that is, the diagonal lines of x + y = K) to divide the status.

Consider enumerating two vertices on the diagonal line X1, Y1, X2, Y2
We strongly use F [X1] [Y1] [X2] [y2] to express the best solution for the two paths to go to points (x1, Y1) and points (X2, Y2) respectively.

In this way, F [X1] [Y1] [X2] [y2] can be composed of F [x1-1] [Y1] [x2-1] [y2], f [x1-1] [Y1] [X2] [y2-1], F [X1] [y1-1] [x2-1] [y2], f [X1] [y1-1] [X2] [y2-1] To get (crazy)

And when the points (x1, Y1) and points (X2, Y2) are adjacent on the diagonal line, it is special because the points (x1-1, Y1) and the points (X2, y2-1) overlap, special considerations (really sad)

 

Although it looks disgusting, I think it is also a big advantage to say that the short stream is so short?

 

#include <cstdio>#include <cstring>#define max(x, y) ((x)>(y) ? (x):(y))const int size=32;int n;int a[size][size];int f[size][size][size][size];inline int getint();inline void putint(int);int main(){while (scanf("%d", &n)!=EOF){for (int i=1;i<=n;i++)for (int j=1;j<=n;j++)a[i][j]=getint();memset(f, 0, sizeof f);f[2][1][1][2]=a[1][1]+a[1][2]+a[2][1];for (int i=4;i<2*n;i++)for (int y1=i<=n+1?1:i-n;y1<i-1 && y1<n;y1++)for (int y2=y1+1;y2<i && y2<=n;y2++){int x1=i-y1, x2=i-y2;if (y2-y1==1) f[x1][y1][x2][y2]=max(max(f[x1][y1-1][x2][y1], f[x2][y1][x2-1][y2]), f[x1][y1-1][x2-1][y2]);else f[x1][y1][x2][y2]=max(max(f[x1-1][y1][x2-1][y2], f[x1-1][y1][x2][y2-1]), max(f[x1][y1-1][x2-1][y2], f[x1][y1-1][x2][y2-1]));f[x1][y1][x2][y2]+=a[x1][y1]+a[x2][y2];}putint(a[n][n]+f[n][n-1][n-1][n]);}return 0;}inline int getint(){register int num=0;register char ch;do ch=getchar(); while (ch<‘0‘ || ch>‘9‘);do num=num*10+ch-‘0‘, ch=getchar(); while (ch>=‘0‘ && ch<=‘9‘);return num;}inline void putint(int num){char stack[15];register int top=0;for ( ;num;num/=10) stack[++top]=num%10+‘0‘;for ( ;top;top--) putchar(stack[top]);putchar(‘\n‘);}

 

[HDU 2686] Matrix