NYOJ 18 The Triangle（简单dp）

The Triangle

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 4

描述

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.

输入

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.

输出

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

样例输入

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

样例输出

30

题意就是求从最上面一层到最下面一层路径数字相加的最大值，所以从下到上更新较为清楚

ac代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
	int a[100][1000];
	int i,j,n;
	scanf("%d",&n);
	memset(a,0,sizeof(a));
	for(i=0;i<n;i++)
	{
		for(j=0;j<=i;j++)
		{
			scanf("%d",&a[i][j]);
		}
	}
	for(i=n-2;i>=0;i--)//从下到上更新
	{
		for(j=0;j<i+1;j++)
		{
			if(a[i+1][j]>a[i+1][j+1])
			a[i][j]+=a[i+1][j];
			else
			a[i][j]+=a[i+1][j+1];
		}
	}
	printf("%d\n",a[0][0]);//最上面的就是最大值
	return 0;
}