有点意思的题目。。。

解题的关键在于， 假设这个数组中负数的个数为N，非负数的个数为M ，N+M=L 。  

然后证明的是， 对于有N个负数的数组， 这N个负数可以任意选择。  这样题目就变成了求解这个数组进行无限次操作，所能得到最小的负数的个数.

然后bfs 直接上。

那个性质很好证明。 不说了

 

C. Yaroslav and Sequence
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Yaroslav has an array, consisting of (2·n - 1) integers. In a single operation Yaroslav can change the sign of exactly n elements in the array. In other words, in one operation Yaroslav can select exactly n array elements, and multiply each of them by -1.

Yaroslav is now wondering: what maximum sum of array elements can be obtained if it is allowed to perform any number of described operations?

Help Yaroslav.

Input

The first line contains an integer n (2 ≤ n ≤ 100). The second line contains (2·n - 1) integers — the array elements. The array elements do not exceed 1000 in their absolute value.

Output

In a single line print the answer to the problem — the maximum sum that Yaroslav can get.

Sample test(s)
input
2
50 50 50
output
150
input
2
-1 -100 -1
output
100
Note

In the first sample you do not need to change anything. The sum of elements equals 150.

In the second sample you need to change the sign of the first two elements. Then we get the sum of the elements equal to 100.

 

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
using namespace std;

int que[10010];
int g[220];
int mark[220];

int main()
{
    int n;
    scanf("%d",&n);
    int cnt=0;
    for(int i=0;i<2*n-1;i++)
    {
        scanf("%d",&g[i]);
        if(g[i]<0) cnt++;
    }
    int qf=1,qd=0;
    que[0]=cnt;
    mark[cnt]=1;
    while(qf>qd)
    {
        int cur=que[qd++];
        for(int i=0;i<=cur;i++)
        {
            int tmp=n-i;
            if(i<=n && tmp<=2*n-1-cur)
            {
                int nwnode=cur-i+tmp;
                if(mark[nwnode]==0)
                {
                    mark[nwnode]=1;
                    que[qf++]=nwnode;
                }
            }
        }
    }
    int tmp;
    for(int i=0;i<=200;i++)
        if(mark[i]==1) 
        {
            tmp=i;
            break;
        }
    int ans=0;
    for(int i=0;i<2*n-1;i++)
    {
        if(g[i]<0)
        {
            g[i]*=-1;
        }
        ans += g[i];
    }
    sort(g,g+2*n-1);
    int sum=0;
    for(int i=0;i<tmp;i++)
        sum+=g[i];
    printf("%d",ans-2*sum);
    return 0;
}