数字三角形

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

题目大意：

给出一个数n,表示这个数字三角形有n层，每层的数字个数逐层递增。从第一行的数开始，每次可以往右下或左下走一格，直到走到最后一行，把沿途经过的数都加起来，如何走使得这个数最大？

思想：

方法一思想：从上往下走——将数字三角形存储在二维数组里，对每一个数组元素a[i][j]它要不加上正上方元素a[i-1][j],要不加上左上方元素a[i-1][j-1] 。即a[i][j]=Max(a[i-1][j],a[i-1][j-1] )+a[i][j].

代码如下：

<span style="font-size:18px;">#include <stdio.h>

int main(void)
{
	int a[101][101]={0};
	int n;
	int i,j,max;
	scanf("%d",&n);
	
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=i;j++)
			scanf("%d",&a[i][j]);
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=i;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];	
		}
	}
	i=n;
	max=a[n][1];
	for(j=2;j<=n;j++)
	{
		if(a[n][j]>max)
			max=a[n][j];
	}
	printf("%d",max);
	return 0;
} </span>

方法二思想：

从下往上走——定义一个二维数组a[n+1][n+1],给最后一行初始化为数字三角形的最后一层，然后从底部往上顶走，直到走到顶部剩一个元素则为最大值。对数组元素a[i][j]（i从n-1开始，i--到1.j从1开始，j++到n)它要不加上正下方元素a[i+1][j],要不加上右下方元素a[i+1][j+1] 。即a[i][j]=Max(a[i+1][j],a[i+1][j+1]  )+a[i][j].

0	0	0	0	0	0
0	30
0	23	21
0	20	13	10
0	7	12	10	10
0	4	5	2	6	5

代码如下：

#include <stdio.h>

int main(void)
{
	int a[101][101]={0};
	int b[101][101]={0};
	int n;
	int i,j,max;
	scanf("%d",&n);
	
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=i;j++)
			scanf("%d",&a[i][j]);
	}
	//初始化 
	i=n;
	for(j=1;j<=n;j++)
		b[i][j]=a[i][j];
	//dp
	for(i=n-1;i>=1;i--)
	{
		for(j=1;j<=n;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",a[1][1]);
	return 0;
}