這道題吧 沒計算時間 因為給了那么多 一算還可以

就直接寫了線段樹，劉汝佳那本模板

然后！poj的g++比C++慢大約500ms。。。。。。。g++tle，C++就過了

Sliding Window

Time Limit: 12000MS	?	Memory Limit: 65536K
Total Submissions: 67576	?	Accepted: 19163
Case 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?	-1	3
?1?[3??-1??-3]?5??3??6??7?	-3	3
?1??3?[-1??-3??5]?3??6??7?	-3	5
?1??3??-1?[-3??5??3]?6??7?	-3	5
?1??3??-1??-3?[5??3??6]?7?	3	6
?1??3??-1??-3??5?[3??6??7]	3	7

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 3 1 3 -1 -3 5 3 6 7

Sample Output

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

Source

POJ Monthly--2006.04.28, Ikki

上代碼 通俗易懂哦~~~~~~看我的就行了

#include<iostream> #include<cstdio> #include<cstring>using namespace std;int Max[5000100]; int Min[5000100]; int pinf = 2000000000; int ninf = -2000000000; int n,k;void build(int l,int r,int cur){if(l == r){scanf("%d",&Max[cur]);Min[cur] = Max[cur];return;}int mid = (l + r)/2;build(l,mid,cur*2);build(mid + 1,r,cur*2 + 1);Min[cur] = min(Min[cur*2],Min[cur*2 + 1]);Max[cur] = max(Max[cur*2],Max[cur*2 + 1]);return; }int ql,qr; int queryMax(int l,int r,int cur){if(ql <= l&&r <= qr){return Max[cur];}int mid = (l + r)/2;int ans = ninf;if(ql <= mid){ans = max(ans,queryMax(l,mid,cur*2));}if(qr > mid){ans = max(ans,queryMax(mid + 1,r,cur*2 +1));}return ans; } int queryMin(int l,int r,int cur){if(ql <= l && r <= qr){return Min[cur];}int mid = (l + r)/2;int ans = pinf;if(ql <= mid){ans = min(ans,queryMin(l,mid,cur*2));}if(qr > mid){ans = min(ans,queryMin(mid + 1,r,cur*2 + 1));}return ans; }int main(){while(scanf("%d%d",&n,&k) != EOF){build(1,n,1);ql = 1;qr = k;printf("%d",queryMin(1,n,1));ql++;qr++; while(qr <= n){printf(" %d",queryMin(1,n,1));ql++;qr++;}printf("\n");ql = 1;qr = k;printf("%d",queryMax(1,n,1));ql++;qr++; while(qr <= n){printf(" %d",queryMax(1,n,1));ql++;qr++;}printf("\n");}return 0; }

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轉載于:https://www.cnblogs.com/xuyanqd/p/9073920.html

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