Optimal Sum CodeForces - 182C

考察:思维
思路:
??比较明显的是要用单调队列,在一段区间内,可以挑选一些数字变化符号.我们求的最大和只有两方式:尽量将负数变正数,尽量将正数变负数.
??在枚举一段\(len\)区间,求最小的\(k\)个负数的绝对值和,然后剩下的数相加.每移动一位,和要做相应变化.所以需要记录左端点属于绝对值还是普通加减.这里利用\(muliset\)存储.

Code

#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int N = 100010;
int n,len,nums[N],k;
LL ans = -1e14;
void solve()
{
	multiset a,b;
	LL sum = 0,res = -1e14;
	for(int i=1,j=1;i<=n;i++)
	{
		if(nums[i]>0)
		{
			b.insert(nums[i]);
			sum+=nums[i];
		}else{
			if(a.size()+1<=k) sum+=abs(nums[i]),a.insert(nums[i]);
			else{
				if(a.size()&&nums[i]<*(--a.end()))
				{
					int x = *(--a.end());
					sum-=abs(x);
					sum+=x;
					b.insert(x);
					a.erase(a.find(x));
					sum+=abs(nums[i]);
					a.insert(nums[i]);
				}else sum+=nums[i],b.insert(nums[i]);
			}
		}
		if(i-j+1==len) res = max(res,sum);
		if(i-j+2>len)
		{
			if(b.count(nums[j]))
			{
				sum-=nums[j];
				b.erase(b.find(nums[j]));
			}else{
				sum-=abs(nums[j]);
				a.erase(a.find(nums[j]));
				if(*(b.begin())<0&&a.size()+1<=k)
				{
					int x = *(b.begin());
					sum-=(LL)x;//多加了b实在的x 
					sum+=(LL)abs(x);
					a.insert(x);
					b.erase(b.begin());
				}
			}
			j++;
		}
	}
	ans = max(res,ans);
}
int main()
{
	scanf("%d%d",&n,&len);
	for(int i=1;i<=n;i++) scanf("%d",&nums[i]);
	scanf("%d",&k);
	solve();
	for(int i=1;i<=n;i++) nums[i] = -nums[i];
	solve();
	printf("%lld\n",ans);
	return 0;
}

原文：https://www.cnblogs.com/newblg/p/15302831.html