思路
求一個最大值,一般就是DP了。初見這個題目像是費用流,而且建邊顯而易見,但是太麻煩,就沒寫……這個題的DP方案也不難,既然是來回,那麼我們可以把一個人拆成兩個人啊,這樣對於“第一個人”枚舉
最後理論上輸出
代碼
#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <queue>
#include <utility>
int nextInt()
{
int num = 0;
char c;
bool flag = false;
while ((c = std::getchar()) == ' ' || c == '\r' || c == '\t' || c == '\n');
if (c == '-')
flag = true;
else
num = c - 48;
while (std::isdigit(c = std::getchar()))
num = num * 10 + c - 48;
return (flag ? -1 : 1) * num;
}
int max4(const int a, const int b, const int c, const int d)
{
int x = std::max(a, b);
if (c > x)
x = c;
if (d > x)
x = d;
return x;
}
const size_t _Siz = 51u;
int f[_Siz][_Siz][_Siz][_Siz] = { 0 };
int a[_Siz][_Siz] = { 0 };
int n, m;
int main(int argc, char **argv)
{
n = nextInt();
m = nextInt();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
a[i][j] = nextInt();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int k = 1; k <= n; k++)
for (int l = j + 1; l <= m; l++)
f[i][j][k][l] = max4(f[i][j - 1][k - 1][l], f[i - 1][j][k][l - 1], f[i][j - 1][k][l - 1], f[i - 1][j][k - 1][l]) + a[i][j] + a[k][l];
std::cout << f[n][m - 1][n - 1][m] << std::endl;
#ifdef __EDWARD_EDIT
std::cin.get();
std::cin.get();
#endif
return 0;
}