P1004 [NOIP2000 提高組] 方格取數

要走兩次因此，考慮一個四維的陣列來實現，然後如果i=k&&j==l的話記得減一次即得到答案。

點選檢視程式碼

#include <iostream>
#include <stack>
#include <cmath>
#include <algorithm>
#include <set>
#include <vector>
#include <climits>
#include <string.h>
#include <map>
#include <queue>
#include <list>
#include <cmath>
#include <iomanip> 
#define int long long 
#define ios ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define lc u<<1
#define rc u<<1|1
#define gcd __gcd
#define double long double
#define endl "\n"
#define INF LLONG_MAX
#define mod  1000000007
#define N 15
const double PI = 3.14159265358979323846;
using namespace std;
int dp[N][N][N][N], a[N][N],n;//dp[1][j][k][l]表示第一次和第二次的最優解
signed main()
{
    ios;
    cin >> n;
    while (1)
    {
        int x, y,z; cin >> x >> y>>z;
        if (x == 0 && y == 0&&z==0)break;
        a[x][y] = z;
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            for (int k = 1; k <= n; k++)
            {
                for (int l = 1; l <= n; l++)
                {
                    dp[i][j][k][l] = max({ dp[i - 1][j][k - 1][l],dp[i][j - 1][k][l - 1],dp[i][j - 1][k - 1][l],dp[i - 1][j][k][l - 1] }) + a[i][j] + a[k][l];
                    if (i == k && j == l)dp[i][j][k][l] -= a[i][j];
                }
            }
        }
    }
    cout << dp[n][n][n][n] << endl;
    return 0;
}