单调队列—— HDU 4193 Non-negative Partial Sums

Non-negative Partial Sums

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1357    Accepted Submission(s): 518

Problem Description

You are given a sequence of n numbers a₀,..., a_n-1. A cyclic shift by k positions (0<=k<=n-1) results in the following sequence: a_k a_k+1,..., a_n-1, a₀, a₁,..., a_k-1. How many of the n cyclic shifts satisfy the condition that the sum of the fi rst i numbers is greater than or equal to zero for all i with 1<=i<=n?

 

Input

Each test case consists of two lines. The fi rst contains the number n (1<=n<=10⁶), the number of integers in the sequence. The second contains n integers a₀,..., a_n-1 (-1000<=a_i<=1000) representing the sequence of numbers. The input will finish with a line containing 0.

 

Output

For each test case, print one line with the number of cyclic shifts of the given sequence which satisfy the condition stated above.

 

Sample Input

3
2 2 1
3
-1 1 1
1
-1
0

 

Sample Output

3
2
0

思路：

 

以一个长度n为5的数串（-10,3,-1,5,4）为例，如下图：

 

为了简化操作，A[]和Sum[]的存储都从1开始，0下标位置的值都赋为0。 

 

A[]存储数字串以及它的一份拷贝，图中绿色方框为原数据（下标为1~n），黄色方框为拷贝（下标为n~2*n）。那么这个数串的一个序列就是图中红色方框圈起来的n个数，这里称红色方框为滑动窗口。

 

Sum[]存储一段数字的和，Sum[i]表示，A[1]~A[i]的和。如上图中Sum[3] = -10 + 3 + -1 = -8。

 

这里设S(i)为：假设滑动窗口的最后一个数的下标为就j，那么j-n+1 <= i <= j，S(i) = A[j-n+1] + A[j-n] + … + A[i]。题中要找满足滑动窗口内所有S(i) >= 0条件的序列，所以只需要找出最小的S(i)，如果它的值都大于等于0的话，该序列就满足条件。使用单调递增队列来寻找最小的S(i)。

 

Sum[i]和S(i)的转化关系为：当前滑动窗口的末尾数的下标为j，S (i) =Sum[i] – Sum[j-n]。

代码：

#include <stdio.h>

#define MAXN 1000000

int sum[2*MAXN+5];	//将长度为n的数串复制为两份存储在sum中，下标从1开始
					//然后，sum[i]存储数串的前i项和
int n;				//数串的长度

int mq[MAXN+5];		//单调递增队列，队列元素为sum数组的下标
int front;			//队首指针
int rear;			//队尾指针，指向队尾的下一个位置

void InitMQ(void)
{
	front = rear = 0;
}

bool IsEmpty(void)
{
	return front == rear;
}

void Push(int i)
{
	while (!IsEmpty() && sum[mq[rear-1]] > sum[i])
	{
		rear--;
	}
	mq[rear++] = i;
}

int Front(void)
{
	return mq[front];
}

void Pop(void)
{
	front++;
}

int main(void)
{
	sum[0] = 0;
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0)
		{
			break;
		}

		int i;
		for (i = 1; i <= n; i++)	//输入数据，并复制为两份
		{
			scanf("%d", &sum[i]);
			sum[i+n] = sum[i];
		}

		for (i = 2; i < n*2; i++)		//计算前i项和
		{
			sum[i] += sum[i-1];
		}

		InitMQ();
		for (i = 1; i < n; i++)
		{
			Push(i);
		}

		int count = 0;
		for (i = n; i < n*2; i++)
		{
			if (Front() + n - 1 < i)	//弹出不在滑动窗口内的数据
			{
				Pop();
			}
			Push(i);
			if (sum[Front()] - sum[i-n] >= 0)	//序列是否满足条件
			{
				count++;
			}
		}

		printf("%d\n", count);
	}
	return 0;
}

/*
5
-3 2 -1 5 3
*/