Project Euler problem 67: find maximum cost path in 100-row triangle

问题

In Project Euler's problem 67 there is a triangle given and it contains 100 rows. For e.g.

        5
      9  6
    4   6  8
  0   7  1   5

I.e. 5 + 9 + 6 + 7 = 27.

Now I have to find the maximum total from top to bottom in given 100 rows triangle.

I was thinking about which data structure should I use so that problem will get solved efficiently.

回答1:

You want to store this as a directed acyclic graph. The nodes are the entries of your triangular array, and there is an arrow from i to j iff j is one row lower and immediately left or right of i.

Now you want to find the maximum weight directed path (sum of the vertex weights) in this graph. In general, this problem is NP-complete, however, as templatetypedef points out, there are efficient algorithms for DAGs; here's one:

algorithm dag-longest-path is
    input: 
         Directed acyclic graph G
    output: 
         Length of the longest path

    length_to = array with |V(G)| elements of type int with default value 0

    for each vertex v in topOrder(G) do
        for each edge (v, w) in E(G) do
            if length_to[w] <= length_to[v] + weight(G,(v,w)) then
                length_to[w] = length_to[v] + weight(G, (v,w))

    return max(length_to[v] for v in V(G))

To make this work, you will need to weight the length of the path to be the size of the target node (since all paths include the source, this is fine).

回答2:

Which language are you using?

A muti-dimensional array is probably the best way to store the values, then depending on the language, there are options for how you store pointers or references to where in the arrays you are.

回答3:

To build off of @Matt Wilson's answer, using a two-dimensional array to hold the numbers would be a fairly simple solution. In your case, you would encode the triangle

      5
    9  6
  4   6  8
0   7  1   5

as the array

[5][ ][ ][ ]
[9][6][ ][ ]
[4][6][8][ ]
[0][7][1][5]

From here, a node at position (i, j) has children at (i + 1, j) and (i + 1, j + 1) and parents at position (i - 1, j) and (i - 1, j - 1), assuming those indices are valid.

The advantage of this approach is that if your triangle has height N, the space required for this approach is N², which is just less than twice the N(N + 1) / 2 space required to actually store the elements. A linked structure like an explicit graph would certainly use more memory than that.

回答4:

I believe this is a Project Euler problem.

It can't be represented with a binary tree. Any kind of graph is overkill as well, though the problem can be solved with a longest path algorithm in directed acyclic graphs. Edit: Nevermind that, it is only useful if the edges are weighted, not the nodes.

A two dimensional vector (e.g. vector<vector<int> >) is more than enough to represent such triangles, and there is a straightforward solution using such representation.

回答5:

#include <iostream>
#include <fstream>
#include <vector>
#include <sstream>

int main() { 
    std::vector<std::vector<int> > lines;
    lines.resize(100);
    std::ifstream input("triangle.txt");

    for (int i = 0; i < 100; i++) {
        for (int j = 0; j < i + 1; j++) {
            std::string number_string;
            input >> number_string;
            std::istringstream temp(number_string);
            int value = 0;
            temp >> value;
            lines[i].push_back(value);
        }
    } 
    std::vector<int> path1;
    path1.resize(100);
    std::vector<int> path2;
    path2.resize(100);

    for (int i = 0;i < 100;i++) 
        path1[i] = lines[99][i];

    for (int i = 98; i >= 0;i--) {  
        for(int j = 0;j < i+1;j++) {
            if(path1[j] > path1[j + 1]){
                path2[j] = path1[j] + lines[i][j];
            } else{
                path2[j] = path1[j + 1] + lines[i][j];
            }
        }
        for (int i = 0;i < 100;i++) 
            path1[i] = path2[i];
    }
    std::cout << path1[0] << std::endl;
}

来源：https://stackoverflow.com/questions/6553702/project-euler-problem-67-find-maximum-cost-path-in-100-row-triangle