POJ - 1163：The Triangle

The Triangle

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题目

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.

输入样例

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

输出样例

30

参考代码1

#include
#include  
using namespace std;

const int maxn=100+10;
int arr[maxn][maxn],maxSum[maxn][maxn];
int n;

int getMaxSum(int x,int y)
{
	if(maxSum[x][y]!=-1)
		return maxSum[x][y];
	if(x==n)
		maxSum[x][y]=arr[x][y];
	else
	{
		int l=getMaxSum(x+1,y);
		int r=getMaxSum(x+1,y+1);
		maxSum[x][y]=max(l,r)+arr[x][y];
	}
	return maxSum[x][y];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=i;j++)
		{
			scanf("%d",&arr[i][j]);
			maxSum[i][j]=-1;
		}
			
	}
	printf("%d\n",getMaxSum(1,1));
}

参考代码2

#include
#include  
using namespace std;

const int maxn=100+10;
int arr[maxn][maxn],maxSum[maxn][maxn];
int n;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=i;j++)
			scanf("%d",&arr[i][j]);

	for(int i=n;i>=1;i--)
	{
		for(int j=1;j<=i;j++)
		{
			if(i==n)
				maxSum[i][j]=arr[i][j];
			else
				maxSum[i][j]=max(maxSum[i+1][j],maxSum[i+1][j+1])+arr[i][j];
		}
	}
	printf("%d\n",maxSum[1][1]);
}

参考代码3

#include
#include  
using namespace std;

const int maxn=100+10;
int arr[maxn][maxn];
int n;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=i;j++)
			scanf("%d",&arr[i][j]);

	int *p=arr[n];
	
	for(int i=n-1;i>=1;i--)
		for(int j=1;j<=i;j++)
			p[j]=max(p[j],p[j+1])+arr[i][j];
			
	printf("%d\n",p[1]);
}