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 a
0,..., 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, a
0, a
1,..., 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<=10
6), the number of integers in the sequence. The second contains n integers a
0,..., 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
*/