hdu 4193 单调队列

题目：

Non-negative Partial Sums

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2849    Accepted Submission(s): 977

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

 

Source

SWERC 2011

给你一个n项的序列，每次可以把序列的首项移动到末尾，显然一共可以构成 n 种序列，问一共有多少种序列满足条件：序列的前 i 项和都大于等于0（i:1~n)。

分析：

环的问题可以通过开双倍数组并将输入数据复制两份接起来来解决

[i,j]的每一个前缀和都大于0等价于从数列起点开始的前缀和在[I,J]部分的最小值都满足比sum(i-1)大

问题转化为高效求[i,j]区间的最小值 即固定区间求最值 联想到单调队列，相当于一个定长的滑块往后滑，滑的过程中通过入队出队操作使得队首值即为最小值

代码：

#include 
using namespace std;
const int maxn=1000010<<1;

int sum[maxn],q[maxn];
int n,cnt,Front,rear;
//队列中维护[1,2n]的下标 使得这些下标对应的前缀和递增 这样队首元素对应的前缀和是最小的
inline void in(int i) {
    while(Front<=rear&&sum[q[rear]]>sum[i]) rear--;//保持前缀和严格递增  或者不降也行
    q[++rear]=i;//插到队尾
}

inline void out(int i,int n) {//删除i使得队列大小为n
    if(q[Front]<=i-n) Front++;//i-n属于[0,n) 从队首移除
}

int main(){//1528MS	9548K
    while (scanf("%d",&n)==1 && n) {
        sum[0]=0;
        cnt=0;
        Front=0;
        rear=-1;
        for(int i=1;i<=n;++i) {
            scanf("%d",&sum[i]);
            sum[i+n]=sum[i];
        }
        for (int i=2;i=0) cnt++;
        }
        printf("%d\n",cnt);
    }
    return 0;
}