dp 方格取数

https://www.luogu.com.cn/problem/P1004

错误示例：跑两遍dp

#include<bits/stdc++.h>

using namespace std;
//    clock_t start, end;
//    start = clock();
//    end = clock();
//    cout << (double) (end - start) / CLOCKS_PER_SEC << endl;
//ios::sync_with_stdio(false);
//cin.tie(nullptr);
#define  int long long
#define lowbit(x) (x&-x)
#define ispow(n) (n & (n - 1)) == 0 //O(1) 判断是否是 2^k（2的k次方）
#define rep(i, x, y) for(int i=(x);i<=(y);++i)
#define dep(i, x, y) for(int i=(x);i>=(y);--i)
#define gcd(a, b) __gcd(a,b)
const long long mod = 1e9 + 7;
const int maxn = 1e3 + 10;

int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n;
int dp[200][200];
int a[200][200];
pair<int,int> pp[200][200];
void solve(){

    for(int i =1;i<=n;i++) {
        for (int j = 1; j <= n; j++) {
            if(dp[i - 1][j]>=dp[i][j - 1]&& i>1) {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + a[i][j];
                pp[i][j].first = i - 1,pp[i][j].second = j;
            }
            else {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + a[i][j];
                pp[i][j].first = i, pp[i][j].second = j - 1;
            }
        }
    }
}

void path(int x, int y)
{
    a[x][y] = 0;
    if(x == 1&& y == 1)
    {
        return ;
    }

    path(pp[x][y].first,pp[x][y].second);
  //  cout<<x<<' '<<y<<endl;
}
signed main() {

    cin>>n;
    int x,y,z;
    while(cin>>x>>y>>z&&x+y+z)
    {
        a[x][y] = z;
    }
    int ans =0;
    solve();
    ans += dp[n][n];
    memset(dp,0,sizeof(dp));
    path(n,n);

    solve();

    cout<< ans+dp[n][n]<<endl;
    return 0;
}

显然，这样做有问题。
举个栗子：
0 2 5
2 6 0
0 6 0
用上面的程序跑出来会有问题。

思考，用四维矩阵，每一次走就有四种走法，就很容易的出转移方程了：

#include<bits/stdc++.h>

using namespace std;
//    clock_t start, end;
//    start = clock();
//    end = clock();
//    cout << (double) (end - start) / CLOCKS_PER_SEC << endl;
//ios::sync_with_stdio(false);
//cin.tie(nullptr);
#define  int long long
#define lowbit(x) (x&-x)
#define ispow(n) (n & (n - 1)) == 0 //O(1) 判断是否是 2^k（2的k次方）
#define rep(i, x, y) for(int i=(x);i<=(y);++i)
#define dep(i, x, y) for(int i=(x);i>=(y);--i)
#define gcd(a, b) __gcd(a,b)
const long long mod = 1e9 + 7;
const int maxn = 1e3 + 10;

int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n;
int dp[20][20][20][20];//第一个人走到 i，j 第二个人走到 k l的最优解
int a[200][200];
pair<int,int> pp[200][200];

signed main() {
    cin>>n;
    int x,y,z;

    while(cin>>x>>y>>z&&x+y+z)
    {
        a[x][y] = z;
    }
    rep(i,1,n)
    {
        rep(j,1,n)
        {
            rep(k,1,n)
            {
                int l = i + j  - k;
                if(l < 0)break;
                    if(i == k&&j == l)
                        dp[i][j][k][l] =
                                max(max(dp[i - 1][j][k][l - 1], dp[i - 1][j][k - 1][l]), max(dp[i][j - 1][k - 1][l],
                                                                                             dp[i][j - 1][k][l - 1])) +a[i][j];
                    else
                        dp[i][j][k][l] = max(max(dp[i - 1][j][k][l - 1], dp[i - 1][j][k - 1][l]),max(dp[i][j - 1][k - 1][l],
                                                                                                 dp[i][j - 1][k][l - 1]))+a[i][j] + a[k][l];
            }
        }
    }
    cout<<dp[n][n][n][n]<<endl;
    return 0;
}