NYOJ 18-The Triangle

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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

这是一道经典的动规问题,从上往下或者从下往上推都可以,我选择的是从上往下,输入完结果就出来了,每行的每个点,都选择上面两个点(能走到自己的只有两个点)中较大的一个,加上自己本身,并记录下来,最后遍历一遍最后一行的数取最大的数就是所求的结果

 
#include<stdio.h>
#include<vector>
using namespace std;
vector<int> tri[102];
int main()
{
	int n;
	scanf("%d", &n);
	int i, j;
	int num;
	scanf("%d", &num);
	tri[0].push_back(num);
	for(i = 1; i < n; i++)
	{
		scanf("%d", &num);
		tri[i].push_back(tri[i - 1][0] + num);
		for(j = 1; j < i; j++)
		{
			scanf("%d", &num);
			tri[i].push_back((tri[i - 1][j - 1] > tri[i - 1][j]? tri[i - 1][j - 1] : tri[i - 1][j]) + num);
		}
		scanf("%d", &num);
		tri[i].push_back(tri[i - 1][j - 1] + num);
		
	}	
	int max = -0x7fffffff;
	i--;
	for(j = 0; j < n; j++)
		if(max < tri[i][j])
			max = tri[i][j];
	printf("%d\n", max);
	return 0;
}