Gym - 100989L dfs

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本文介绍了一个数学问题：如何通过最少的符号更改使一个由加减运算组成的等式成立。利用深度优先搜索（DFS）算法，文章提供了一种有效的解决方案，并附带了完整的C++实现代码。

Time limit

1000 ms

Memory limit

262144 kB

题目描述：

AbdelKader enjoys math. He feels very frustrated whenever he sees an incorrect equation and so he tries to make it correct as quickly as possible!

Given an equation of the form: A1 o A2 o A3 o ... o An = 0, where o is either + or -. Your task is to help AbdelKader find the minimum number of changes to the operators + and -, such that the equation becomes correct.

You are allowed to replace any number of pluses with minuses, and any number of minuses with pluses.

Input

The first line of input contains an integer N (2 ≤ N ≤ 20), the number of terms in the equation.

The second line contains N integers separated by a plus + or a minus -, each value is between 1 and 108.

Values and operators are separated by a single space.

Output

If it is impossible to make the equation correct by replacing operators, print - 1, otherwise print the minimum number of needed changes.

Examples

Input

7
1 + 1 - 4 - 4 - 4 - 2 - 2

Output

Input

3
5 + 3 - 7

Output

-1

简单dfs

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int ans=30,flag=0,al[25],n;

char bl;
void dfs(int x,int y,int dfst,int f)

// x(深度) y(是否改变符号(1为不变)(-1为改变)) dfst(为dfs到x深度时的和) f(记录符号改变的次数)

//因为只需要得到最后一组的数据所以使用dfst和f来记录前x个数总和和前x次符号改变次数的总和

//不需要使用数组来生成线段树记录每一次dfs的结果
{
    dfst=al[x]*y+dfst;
    if(y==-1)
    {
        f=f+1;
    }
    else
    {
        f=f;
    }
    if(x==n)
    {
        if(dfst==0)
        {
            flag=1;
            ans=min(ans,f);
        }
        return ;
    }
    dfs(x+1,1,dfst,f);//符号不变
    dfs(x+1,-1,dfst,f);//符号改变
}
int main()
{
    cin>>n;
    for(int n1=1;n1<=n;n1++)
    {
        if(n1==1)
        {
            cin>>al[n1];
        }
        else
        {
            getchar();
            cin>>bl;
            getchar();
            cin>>al[n1];
            if(bl=='-')
            {
                al[n1]=-al[n1];
            }
        }
    }
    dfs(1,1,0,0);
    if(flag==1)//flag判断能否得到和为0
    {
        cout<<ans<<endl;
    }
    else
    {
        cout<<-1<<endl;
    }
}