动态规划经典入门题目

作为动态规划的经典入门题目，就是这道题啦。

7
3   8
8   1   0
2   7   4   4
4   5   2   6   5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

Sample Output

30

Hint

嗯，这道题的思路就非常简单了，给出关键的递推公式吧：

getmax[i][j]=max(getmax[i+1][j],getmax[i+1][j+1])+graph[i][j]。（从数字金字塔的底部向上递推）

下面是AC代码

#include
#include
using namespace std;
const int maxn=100;
int graph[maxn][maxn];
int getmax[maxn][maxn];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i=0;i--)
	{
		for(int j=0;j<=i;j++)
		{
			getmax[i][j]=max(getmax[i+1][j],getmax[i+1][j+1])+graph[i][j];
		}
		
	}
	printf("%d\n",getmax[0][0]);
	return 0;
	
}