字符串中插入加号（dp）动态规划

Posted 2020-08-09 mahaitao617

题目：长度为m的字符串插入n个加号求最小和。例如string str="123456",n=2;输出12+34+56的和为102，同时输出2 4，也就是加号位置。下面为实现思路：

实现过程主要就是区间dp的思想，其中状态转移方程为dp[i][j] = min(dp[k][j - 1] + getnum(str.substr(k, i-k)), dp[i][j]);//其中dp[i][j]表示1-i中使用了j个加号，在这里我用struct主要就是记录加号的位置。

#include<iostream>  
#include<ctime>
#include<vector>
#include<string>
#include<algorithm>
#include<sstream>
using namespace std;
//把字符串转化成对应的整数例如字符串"123"的getnum值就是123
int getnum(string str)
{
	if (str.size() <= 0)return 0;
	int sum = 0;
	for (int i = 0; i < str.size(); i++)
	{
		sum = 10 * sum + (str[i] - ‘0‘);
	}
	return sum;
}
//把字符串中每个数字分别相加例如"123"的getnum1就是6
int getnum1(string str)
{
	int sum = 0;
	for (int i = 0; i < str.size(); i++)
	{
		sum = sum + (str[i] - ‘0‘);
	}
	return sum;
}
struct node
{
	int value;
	int k;
};
int main()
{

	string str = "1234567891";
	int n = 3;
	int m = str.size();
	vector<vector<node>>dp(m + 1, vector<node>(n+1));
	for (int i = 0; i < m+1; i++)
	{
		for (int j = 0; j < n+1; j++)
		{
			dp[i][j].value = 9999999;
			dp[i][j].k = 0;
		}
	}
	vector<int>recordpos;//记录加号的位置
	dp[0][0].value = 0;
	int index = 0;
	for (int i = 0; i <=m; i++)
	{
		dp[i][0].value = getnum(str.substr(0,i));
	}
	for (int j = 0; j <=n; j++)
	{
		dp[0][j].value = 0;
	}
	for (int i = 1; i < m+1; i++)
	{
		for (int j =1; j< n+1; j++)
		{
			if (j >= i - 1)dp[i][j].value = getnum1(str.substr(0, i));
			else
			{
				for (int k = 1; k <=i; k++)
				{
					//dp[i][j] = min(dp[k][j - 1] + getnum(str.substr(k, i-k)), dp[i][j]);//其中dp[i][j]表示1-i中使用了j个加号
					if (dp[i][j].value>dp[k][j - 1].value + getnum(str.substr(k, i - k)))
					{
						dp[i][j].value = dp[k][j - 1].value + getnum(str.substr(k, i - k));
						//if (i==m)recordpos.push_back(k);
						dp[i][j].k = k;
					}
				}
				
			}
		}
	}
	cout << dp[m][n].value << endl;
	cout << dp[m][n].k << endl;
	cout << dp[4][1].k << endl;
	int k = dp[m][n].k;
	int y = n;
	while (k>0 && n>0)
	{
		recordpos.push_back(k);
		y--;
		m = k;
		k = dp[m][y].k;
	}
	for (int i = 0; i < recordpos.size(); i++)
	{
		cout << recordpos[i] << " ";
	}
	cout << endl;
	cin.get();
	return 0;
}