POJ 1163 The Triangle (DP)

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题目链接。这个题目很简单，于是很快就1次过了。DP问题。

直接给出状态转移方程：

dpRes[i][j] = max(dpRes[i-1][j-1], dpRes[i-1][j])+A[i][j]

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dpRes[i][j]表示前i行第j个数包含在内的最大和，A[i][j]为题目给的三角形（矩阵）。扫描对于最后一行得出最大值即为题目所求。为了节省时间，在输入完成一行后就直接计算出这一行当dpRes[i][j]，(for j=1 to i)。我的代码如下：

   1: #include <iostream>

   2: using namespace std;

   3: const int ROW_SIZE = 101;

   4: int dpRes[ROW_SIZE][ROW_SIZE];

   5: int triangle[ROW_SIZE][ROW_SIZE];

6:

   7: int main()

   8: {

   9:     int row=0;

  10:     cin >> row;

  11:     if (row<=0 || row>=ROW_SIZE)

  12:         return -1;

  13:     for (int i=1; i<=row; i++)

  14:     {

  15:         for (int j=1; j<=i; j++)

  16:             cin >> triangle[i][j];

  17:         for (int j=1; j<=i; j++)

  18:         {

  19:             int temp = dpRes[i-1][j-1]>dpRes[i-1][j] ? dpRes[i-1][j-1] : dpRes[i-1][j];

  20:             dpRes[i][j] = temp + triangle[i][j];

  21:         }

  22:     }

  23:     int max=0;

  24:     for (int j=1; j<=row; j++)

  25:         max = max<dpRes[row][j] ? dpRes[row][j] : max;

  26:     cout << max << endl;

  27:     return 0;

  28: }