数字三角形

测试样例：

5

7

3 8

8 1 0
2 7 4 4
4 5 2 6 5

这种方法的时间复杂度为O(2n)，需要继续优化 
 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 constexpr int MAX = 101;
 5 int D[MAX][MAX];
 6 int n;
 7 int Maxsum(int i, int j)
 8 {
 9     if (i == n)
10         return D[i][j];
11     else
12         return max(Maxsum(i + 1, j), Maxsum(i + 1, j + 1)) + D[i][j];
13 }
14 int main(void)
15 {
16     int i, j;
17     cin >> n;
18     for (i = 1; i<=n;++i)
19     {
20         for (j = 1; j <= i; ++j)
21         {
22             cin >> D[i][j];
23         }
24     }
25     cout << Maxsum(1, 1) << endl;
26     return 0;
27 }

 这种方法的时间复杂度为O(n2)，需要继续优化
 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 constexpr int MAX = 101;
 5 int D[MAX][MAX];
 6 int maxsum[MAX][MAX];
 7 int n;
 8 int Maxsum(int i, int j)
 9 {
10     if (maxsum[i][j] != -1)
11         return maxsum[i][j];
12     if (i == n)
13         maxsum[i][j] = D[i][j];
14     else
15     {
16         maxsum[i][j] = max(Maxsum(i+1,j), Maxsum(i+1,j+1))+ D[i][j];
17     }
18     return maxsum[i][j];
19 }
20 int main(void)
21 {
22     int i, j;
23     cin >> n;
24     for (i = 1; i<=n;++i)
25     {
26         for (j = 1; j <= i; ++j)
27         {
28             cin >> D[i][j];
29             maxsum[i][j] = -1;
30         }
31     }
32     cout << Maxsum(1, 1) << endl;
33     return 0;
34 }

 这种方法与上一种相同，时间复杂度为O(n2)，需要继续优化
 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 constexpr int MAX = 101;
 5 int D[MAX][MAX];
 6 int maxsum[MAX][MAX];
 7 int n;
 8 int main(void)
 9 {
10     int i, j;
11     cin >> n;
12     for (i = 1; i<=n;++i)
13     {
14         for (j = 1; j <= i; ++j)
15         {
16             cin >> D[i][j];
17         }
18     }
19     for (j = 1; j <= n; ++j)
20     {
21         maxsum[n][j] = D[n][j];
22     }
23     for (i = n - 1; i >= 1; --i)
24     {
25         for (j = 1; j <= i; ++j)
26         {
27             maxsum[i][j] = max(maxsum[i + 1][j], maxsum[i + 1][j + 1]) + D[i][j];
28         }
29     }
30 
31     cout <<maxsum[1][1]<< endl;
32     return 0;
33 }

 这种方法与上一种相同，时间复杂度为O（n2），但空间复杂度为O(n)
 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 constexpr int MAX = 101;
 5 int D[MAX][MAX];
 6 int *maxsum;
 7 int n;
 8 int main(void)
 9 {
10     int i, j;
11     cin >> n;
12     for (i = 1; i<=n;++i)
13     {
14         for (j = 1; j <= i; ++j)
15         {
16             cin >> D[i][j];
17         }
18     }
19     maxsum = D[n];
20     for (i = n - 1; i >= 1; --i)
21     {
22         for (j = 1; j <= i; ++j)
23         {
24             maxsum[j] = max(maxsum[j], maxsum[j + 1]) + D[i][j];
25         }
26     }
27     cout <<maxsum[1]<< endl;
28     return 0;
29 }