[POJ] The Triangle

最新推荐文章于 2019-05-23 17:41:08 发布

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最新推荐文章于 2019-05-23 17:41:08 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/KennyRom/p/6477084.html

本文介绍了一个经典的动态规划问题——求解三角形路径中的最高数值和。通过递归地从底部向上更新每个节点的最大可能值，最终得到顶部节点的值即为所求的最大路径和。

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The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47278		Accepted: 28608

Description

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

Sample Output

//利用动态规划解决问题
#include<iostream>
using namespace std;

int a[101][101]; //二维数组储存三角形中的数

int main()
{
	int n;
	cin>>n;

	for(int i=0;i<n;i++)
		for(int j=0;j<=i;j++)
			cin>>a[i][j];

	for(int i=n-2;i>=0;i--)
		for(int j=0;j<=i;j++)
			a[i][j]+=(a[i+1][j]>a[i+1][j+1]?a[i+1][j]:a[i+1][j+1]);
    //动态规划决策，从后往上加，加到最后最大值为顶端的值

	cout<<a[0][0]<<endl;

	return 0;
}

转载于:https://www.cnblogs.com/KennyRom/p/6477084.html