hdu 4193 单调队列

来源：互联网发布：美国历年财政支出数据编辑：程序博客网时间：2024/06/14 02:04

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 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

32 2 13-1 1 11-10

Sample Output

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 <bits/stdc++.h>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(){//1528MS9548K    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<n+n;++i) sum[i]+=sum[i-1];        for(int i=1;i<n;++i) in(i);        //现在队列里面有n-1个元素 后面每进一个就要出一个        for(int i=n;i<n+n;++i){            in(i);            out(i,n);            if(sum[q[Front]]-sum[i-n]>=0) cnt++;        }        printf("%d\n",cnt);    }    return 0;}

阅读全文

0 0