poj 2823 Sliding Window(单调队列模板)

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介绍单调队列不错的博客:点击打开链接


还有这题输出用printf G++交会T,C++就能过.....而用cout两个都能过..好坑....


代码:

#include<cstdio>using namespace std;const int maxn = 1e6+5;int n, k, ansMin[maxn], ansMax[maxn], queMax[maxn], queMin[maxn], val[maxn];int main(void){    while(~scanf("%d%d", &n, &k))    {        int headMin = 0, tailMin = 0, headMax = 0, tailMax = 0;        for(int i = 0; i < n; i++)        {            while(headMin < tailMin && queMin[headMin] <= i-k) headMin++;            while(headMax < tailMax && queMax[headMax] <= i-k) headMax++;            scanf("%d", &val[i]);            while(headMin < tailMin && val[queMin[tailMin-1]] >= val[i]) tailMin--; tailMin++;            while(headMax < tailMax && val[queMax[tailMax-1]] <= val[i]) tailMax--; tailMax++;            queMin[tailMin-1] = queMax[tailMax-1] = i;            ansMin[i] = val[queMin[headMin]];            ansMax[i] = val[queMax[headMax]];        }        for(int i = k-1; i < n; i++) printf("%d%c", ansMin[i], i==n-1 ? '\n' : ' ');        for(int i = k-1; i < n; i++) printf("%d%c", ansMax[i], i==n-1 ? '\n' : ' ');    }    return 0;}



Sliding Window
Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 56208 Accepted: 16159Case Time Limit: 5000MS

Description

An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example: 
The array is [1 3 -1 -3 5 3 6 7], and k is 3.Window positionMinimum valueMaximum value[1  3  -1] -3  5  3  6  7 -13 1 [3  -1  -3] 5  3  6  7 -33 1  3 [-1  -3  5] 3  6  7 -35 1  3  -1 [-3  5  3] 6  7 -35 1  3  -1  -3 [5  3  6] 7 36 1  3  -1  -3  5 [3  6  7]37

Your task is to determine the maximum and minimum values in the sliding window at each position. 

Input

The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line. 

Output

There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values. 

Sample Input

8 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 7

Source

POJ Monthly--2006.04.28, Ikki


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