Codeforces 1447B - B. Numbers Box Acwing 3763.数字矩阵 每日一题7.12

题目

        Codeforces链接Problem - 1447B - Codeforces

        AcWing链接3763. 数字矩阵 - AcWing题库

思路

        如果一个n*m的矩阵里面有许多分散的负数，我们可以两两交换使得负数接近，那么我们就看是否有两个两个配对，也就是负数是奇数还是偶数，这时候我们把0也算上，把0看作一个负数，对答案不影响，而且还更好，减去一个0。

代码

        

#include<stack>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<deque>
#include<vector>
#include<iostream>
#include<map>
#include<set>
#include<iomanip>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
using namespace std;
ll GCD(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int inf=0x3f3f3f3f;
int minn,n,m,x,sum,num,T;
signed main() 
{
    #ifndef ONLINE_JUDGE
		    freopen("IO\\in.txt","r",stdin);
		    freopen("IO\\out.txt","w",stdout);
        #endif
    IOS
    cin>>T;
    while (T--)
    {
        cin>>n>>m;
        minn=inf;
        num=0;
        sum=0;
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
            {
                cin>>x;
                sum+=abs(x);
                if (x<=0) num++;
                minn=min(abs(x),minn);
            }
        if (num%2==1) cout<<sum-minn*2<<endl;
        else cout<<sum<<endl;
    }
    return 0;
}