dfs與dp算法之關系與經典入門例題

目錄

• 聲明
• dfs與dp的關系
• 經典例題-數字三角形 - POJ 1163
  • 題目
  • dfs思路
    • 解題思路
    • 具體代碼
  • dp思路
    • 解題思路
    • 具體代碼

聲明

本文不介紹dfs、dp算法的基礎思路，有想了解的可以自己找資源學習。

本文適合於剛剛接觸dfs和dp算法的人，發現兩種算法間的內在聯系。

本人算法之路走之甚短，如果理解出現問題歡迎大家的指正，我會分享基於我目前理解到的算法思想。

dfs與dp的關系

很多情況下，dfs和dp兩種解題方法的思路都是很相似的，這兩種算法在一定程度上是可以互相轉化的。

想到dfs也就常常會想到dp，當然在一些特定的適用場合下除外。

dp主要運用了預處理的思想，而dfs則是類似於白手起家，一步步探索。一般來講，能夠預處理要好些，好比戰爭前做好准備。

dfs和dp都是十分重要的基礎算法，在各個競賽中都有涉及，務必精通。

經典例題-數字三角形 - POJ 1163

題目

The Triangle

Description

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

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

Sample Output

30

dfs思路

---

解題思路

自頂向下，將每種路徑都走一遍。

通過迭代計算到最后一層，記錄最后一層的所有值。

最后一層中的最大值即為所求。

具體代碼

#include <iostream>
// use vector vessel to write down the final level
#include <vector>
// use 'sort' method in <algorithm>
#include <algorithm>
// use 'greater<T>' functional template in <functional>
#include <functional>
using namespace std;
// the maximum of the triangle ordinal
const int max_ordinal = 100;
// the depth
int num_of_rows;
// save data
int data[max_ordinal][max_ordinal];
// save the data of the final level
vector<int> ans;

void dfs(int level, int sum, int column)
{
  // avoid multi calculate
  int current_sum = sum+data[level][column];
  // save data which was in final level
  if(level+1 == num_of_rows)
  {
    ans.push_back(current_sum);
    return;
  }
  // binary tree
  dfs(level+1, current_sum, column);
  dfs(level+1, current_sum, column+1);
}

int main()
{
  cin >> num_of_rows;
  for(int i = 0; i < num_of_rows; i++)
    for(int j = 0; j <= i; j++)
      scanf("%d", &data[i][j]);
  dfs(0, 0, 0);
  cout << "output data:" << endl;
  sort(ans.begin(), ans.end(), greater<int>());
  for(int i = 0; i < ans.size(); i++)
  {
    cout << ans[i] << "\t";
    if(!((i+1) % 5)) cout << endl;
  }
  cout << endl;

  return 0;
}

dp思路

---

解題思路

dfs的思路是從上到下，而dp的思路是：從第二層開始，每下去一次往回看一下並取上一層相鄰兩個大的那個。

具體代碼

#include <iostream>
// use 'max' method and 'sort' method
#include <algorithm>
// use 'greater<T>' functional template
#include <functional>
using namespace std;
// same as DFS
const int max_ordinal = 100;
int num_of_rows;
int data[max_ordinal][max_ordinal];
// the array of the dp method
int dp[max_ordinal][max_ordinal];

int main()
{
	cin >> num_of_rows;
	for(int i = 0; i < num_of_rows; i++)
		for(int j = 0; j<= i; j++)
			scanf("%d", &data[i][j]);
	// dp now
	dp[0][0] = data[0][0];
	for(int i = 1; i < num_of_rows; i++)
	{
		for(int j = 0; j <= i; j++)
		{
			if(j-1 >= 0)
			{
				dp[i][j] = data[i][j] + max(dp[i-1][j], dp[i-1][j-1]);
			} else {
				dp[i][j] = data[i][j] + dp[i-1][j];
			}
		}
	}
  // calling 'sort' method
	sort(dp[num_of_rows-1], &dp[num_of_rows-1][num_of_rows], greater<int>());
	for(int i = 0; i < num_of_rows; i++)
		cout << dp[num_of_rows-1][i] << " ";
	cout << endl;
	cout << "answer is: ";
	cout << dp[num_of_rows-1][0] << endl;

	return 0;
}