牛客小白月赛1 G あなたの蛙は旅⽴っています【图存储】【DP】

题目链接：https://www.nowcoder.com/acm/contest/85/G

思路：

DP 空间可以优化成一维的， 用一维数组的 0 号单元保存左斜对角的值即可。

存图这里真不好理解 = =

AC 代码：

 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <iostream>
 #include <map>
 #include <queue>
 #include <set>
 #include <vector>

 using namespace std;

 #define max3(x, y, z) max(max((x), (y)), (z))
 #define min3(x, y, z) min(mix((x), (y)), (z))
 #define pb push_back
 #define ppb pop_back
 #define mk make_pair

 #define debug_l(a) cout << #a << " " << (a) << endl
 #define debug_b(a) cout << #a << " " << (a) << " "
 #define testin(filename) freopen((filename) ,"r",stdin)
 #define testout(filename) freopen((filename) ,"w",stdout)

 typedef long long ll;
 typedef unsigned long long ull;

 const double PI  = 3.14159265358979323846264338327;
 const double E   = exp();
 const double eps = 1e-;

 const int INF    = 0x3f3f3f3f;
 const int NINF   = 0xc0c0c0c0;
 const int maxn   = 3e3 + ;
 const int MOD    = 1e9 + ;

 int Map[maxn][maxn], Cur[maxn], dp[maxn][maxn];
 int main()
 {
     //testin("data1.in");
     int n;
     scanf("%d", &n);
     memset(Map, NINF, sizeof(Map));
     memset(Cur, , sizeof(Cur));
     memset(dp, , sizeof(dp));
     int vis =  + ( * (n - ));
     int cur =  * n - ;
     int i, j;

     for (int i = ; i < vis - n; i++) {
         if (i <= i % n)
             Cur[i] = i % n + ;
         else {
             if (n & )
                 Cur[i] = (i & ) ? n -  : n;
             else
                 Cur[i] = (i & ) ? n : n - ;
         }
     }
     for (int i = vis - ; i >= vis - n; i--)
         Cur[i] = vis - i;

     vector <int> v[vis];
     int temp;
     for (i = ; i < vis; i++)
     {
         for (j = ; j < Cur[i]; j++)
         {
             scanf("%d", &temp);
             v[i].push_back(temp);
         }
     }

     int len = cur /  + ;
     int flag = ;
     for (i = , j = n; i < len; i++, j++)
     {
         for (int l = , k = flag; l < j; l++, k++)
         {
             Map[i][l] = v[k][v[k].size() - ];
             v[k].pop_back();
             if (v[k].size() == )
                 flag++;
         }
     }

     for (j -= ; i < cur; i++, j--)
     {
         for (int l = (cur - j), k = flag; l < cur; l++, k++)
         {
             Map[i][l] = v[k][v[k].size() - ];
             v[k].pop_back();
             if (v[k].size() == )
                 flag++;
         }
     }

     dp[][] = Map[][];
     for (int i = ; i < cur; i++) {
         dp[][i] = Map[][i] + dp[][i - ];
         dp[i][] = Map[i][] + dp[i - ][];
     }

     for (int i = ; i < cur; i++)
         for (int j = ; j < cur; j++)
             dp[i][j] = Map[i][j] + max3(dp[i - ][j - ], dp[i - ][j], dp[i][j - ]);

     cout << dp[cur - ][cur - ] << endl;
     return ;
 }

参考原文：https://blog.csdn.net/Dup4plz/article/details/79639771