思路
求一个最大值,一般就是DP了。初见这个题目像是费用流,而且建边显而易见,但是太麻烦,就没写……这个题的DP方案也不难,既然是来回,那么我们可以把一个人拆成两个人啊,这样对于“第一个人”枚举 i 行, fi,j 表示到第 j 个格子时的最佳方案,对于“第二个人”枚举 k 行 fk,l 同理。这样我们就凑出了一个四维DP:
fi,j,k,l=max(fi−1,j,k−1,l,fi,j−1,k−1,l,fi−1,j,k,l−1,fi,j−1,k,l−1)+ai,j+ak,l
最后理论上输出
fn,m,n,m ,实际上我们可以发现在计算过程中只算到了
fn,m−1,n−1,m ,所以输出后者就可以啦。
代码
#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;
}