Minimum path sum--LeetCode-优快云博客

本文探讨了如何使用动态规划解决寻找从网格左上角到右下角的最小路径和问题。通过初始化边界条件和迭代更新dp数组，最终得出最优路径。

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题目：

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

思路：

这是一个典型的动态规划，我们需要一个变量记录，dp[i][j]表示到达{i,j}的最短距离，如果不考虑边界处理的话，那么dp[i][j] = min(dp[i][j], min(vec[i][j]+dp[i][j-1],vec[i][j]+dp[i-1][j]))

然后在初始化时再考虑一下边界处理。

#include <iostream>
#include <string>
#include <vector>
#include <limits>
using namespace std;
/*
 Given a m x n grid filled with non-negative numbers, 
 find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.
*/

/*
格子取数 
*/

int MinPahtSum(vector<vector<int> >& vec)
{
	vector<vector<int> > dp(vec.size());
	int i,j;
	for(i=0;i<vec.size();i++)
		dp[i].assign(vec[i].size(),numeric_limits<int>::max());
	dp[0][0] = vec[0][0];
	for(i=1;i<vec.size();i++)
		dp[i][0] = vec[i][0]+dp[i-1][0];
	for(j=1;j<vec[0].size();j++)
		dp[0][j] = vec[0][j] + dp[0][j-1];
	
	int tmp;
	for(i=1;i<vec.size();i++)
	{
		for(j=1;j<vec[0].size();j++)
		{
			tmp = min(vec[i][j]+dp[i][j-1],vec[i][j]+dp[i-1][j]);
			dp[i][j] = min(dp[i][j],tmp);
		}
			
			
	}
	
	 
	return dp[vec.size()-1][vec[0].size()-1];
}
int main()
{
	vector<vector<int> > vec(3);
	int i,j;
	int array[]={2,4,3,7};
	int array1[]={5,3,2,1};
	int array2[]={4,8,6,2};
	vec[0].assign(array,array+4);
	vec[1].assign(array1,array1+4);
	vec[2].assign(array2,array2+4);
	 
	cout<<MinPahtSum(vec)<<endl;
	return 0;
}