Least Prefix Sum（贪心）

题目链接

题目描述：

Baltic, a famous chess player who is also a mathematician, has an array \(a_1,a_2,…,a_n\), and he can perform the following operation several (possibly \(0\)) times:

• Choose some index \(i (1≤i≤n)\);

• multiply \(a_i\) with \(−1\), that is, set \(a_i:=−a_i\).

Baltic's favorite number is \(m\), and he wants \(a_1+a_2+⋯+a_m\) to be the smallest of all non-empty prefix sums. More formally, for each \(k=1,2,…,n\) it should hold that

Please note that multiple smallest prefix sums may exist and that it is only required that \(a_1+a_2+⋯+a_m\) is one of them.

Help Baltic find the minimum number of operations required to make \(a_1+a_2+⋯+a_m\) the least of all prefix sums. It can be shown that a valid sequence of operations always exists.

输入描述：

Each test contains multiple test cases. The first line contains the number of test cases \(t (1≤t≤10000).\) The description of the test cases follows.

The first line of each test case contains two integers \(n\) and \(m (1≤m≤n≤2⋅10^5)\) — the size of Baltic's array and his favorite number.

The second line contains \(n\) integers \(a_1,a_2,…,a_n (−10^9≤ai≤10^9)\) — the array.

It is guaranteed that the sum of \(n\) over all test cases does not exceed \(2⋅10^5\).

输出描述：

For each test case, print a single integer — the minimum number of required operations.

样例：

input:

6
4 3
-1 -2 -3 -4
4 3
1 2 3 4
1 1
1
5 5
-2 3 -5 1 -20
5 2
-2 3 -5 -5 -20
10 4
345875723 -48 384678321 -375635768 -35867853 -35863586 -358683842 -81725678 38576
-357865873

output:

1
1
0
0
3
4

AC代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int T = 1;
int n, m;
LL a[N];

// 要让 m 的前缀和最小，需要让m + 1 ~ n的值都大于0，让m ~ 2的值都小于0
// 要让改变次数最少，则需要找到m + 1 ~ n中小于0的最小值，m ~ 2的大于0的最大值
void solve()
{
	scanf("%d%d", &n, &m);

	for(int i = 1; i <= n; i ++)
		scanf("%lld",&a[i]);

	priority_queue<LL, vector<LL>, greater<LL>> q; // 小根堆
	priority_queue<LL, vector<LL>, less<LL>> p; // 大根堆

	LL sum = 0;
	int ans = 0; // 要用到的次数

	// 遍历m + 1 到 n
	for(int i = m + 1; i <= n; i ++)
	{	
		// m + 1 到 n 的前缀和
		sum += a[i];

		// 如果小于零则存进堆中
		if(a[i] < 0)
			q.push(a[i]);

		// 和小于零则必须要将其变为正的不然m + 1 ~ n的和加上前面的数的和必然小于m的前缀和
		while(sum < 0)
		{
			int t = q.top();
			q.pop();

			sum -= 2 * t;

			ans ++;
		}
	}

	sum = 0;

	// 从后往前遍历，要用最少的次数
	for(int i = m; i >= 2; i --)
	{
		sum += a[i];

		// 如果大于零则存入堆中
		if(a[i] > 0)
			p.push(a[i]);

		// 和大于零则必须将其变为负的不然m加上前面的和必然大于前面的某个数的前缀和
		while(sum > 0)
		{
			int t = p.top();
			p.pop();

			sum -= 2 * t;

			ans ++;
		}
	}

	printf("%d\n", ans);
}

int main()
{
	// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

	scanf("%d", &T);

	while(T --)
		solve();

	return 0;
}