Suffix expression (greed + thinking)

analysis ： First, if the given symbols are all positive signs , Then you can only add up all the numbers as the result , If the given symbol contains a symbol , We can make the following changes ：

Ax-（Ai-Aj+Ak）=Ax-Ai+Aj-Ak, If the plus sign is placed in brackets , Then the plus sign can become a minus sign , The plus sign is placed outside the brackets , The same goes for the minus sign , Put it in brackets and turn it into a plus sign , And the minus sign is still placed outside the brackets , But because there is a minus sign in front of the bracket , So the first number in brackets must be preceded by a minus sign , The number to the left of the bracket must be a plus sign , So the number of numbers we can add is 1~n+m, According to the greedy thought, it is not difficult to figure out that the number on the left of the bracket should take the maximum , And the first number in brackets is taken as a minus , Therefore, the minimum value should be taken , That's the idea , Now let's look at the code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=2e5+10;
long long a[N];
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n+m+1;i++)
		scanf("%lld",&a[i]);
	long long ans=0;
	if(!m)
	{
		for(int i=1;i<=n+m+1;i++)
			ans+=a[i];
	}
	else
	{
		sort(a+1,a+n+m+2);
		ans=a[n+m+1]-a[1];
		for(int i=2;i<=n+m;i++)
			ans+=abs(a[i]);
	}
	printf("%lld",ans);
	return 0;
}