• codechef Sums in a Triangle题解


    Let's consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the largest of the sums of numbers that appear on the paths starting from the top towards the base, so that:

    • on each path the next number is located on the row below, more precisely either directly below or below and one place to the right;
    • the number of rows is strictly positive, but less than 100
    • all numbers are positive integers between O and 99.

    Input

    In the first line integer n - the number of test cases (equal to about 1000).
    Then n test cases follow. Each test case starts with the number of lines which is followed by their content.

    Output

    For each test case write the determined value in a separate line.

    Example

    Input:
    2
    3
    1
    2 1
    1 2 3
    4 
    1 
    1 2 
    4 1 2
    2 3 1 1 
    
    Output:
    5
    9

    使用动态规划法解决本题。

    和一题leetcode题一样,从底往上查找路径。


    #include <vector>
    #include <string>
    #include <algorithm>
    #include <stdio.h>
    #include <iostream>
    using namespace std;
    
    int triPath(vector<vector<int> > &tri)
    {
    	if (tri.empty()) return 0;
    	vector<int> path(tri.back());
    	for (int i = (int)tri.size() - 2; i >= 0 ; i--)//unsigned做减法会溢出!!!
    	{
    		for (int j = 0; j < (int)tri[i].size(); j++)
    		{
    			path[j] = max(path[j], path[j+1]) + tri[i][j];
    		}
    	}
    	return path.front();
    }
    
    int SumsinATriangle()
    {
    	int T, n;
    	scanf("%d", &T);
    	while (T--)
    	{
    		scanf("%d", &n);
    		vector<vector<int> > tri;
    		for (int i = 1; i <= n; i++)
    		{
    			vector<int> tmp(i);
    			for (int j = 0; j < i; j++)
    			{
    				scanf("%d", &tmp[j]);
    			}
    			tri.push_back(tmp);
    		}
    		printf("%d
    ", triPath(tri));
    	}
    	return 0;
    }




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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/3941292.html
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