Sliding Window
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Time Limit: 12000MS |
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Memory Limit: 65536K |
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Total Submissions: 7213 |
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Accepted: 1859 |
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Case Time Limit: 5000MS |
Description
An array of size
n ≤ 10
6 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 position | Minimum value | Maximum 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
這題讓我想到了 幾個(gè)月前的浙大月賽題的區(qū)間最大最小值，用隊(duì)列維護(hù)的方法還是不會(huì)
用線段樹AC，有點(diǎn)郁悶，我的半吊子線段樹啊?? 居然10s才ac，不過幸好服務(wù)器沒掛。。。。。。
代碼如下
?
#include<stdio.h>
#include<string.h>
#define?MAX?1001010
struct?node
{
????int?l,r;
????int?m;
};
node?Max_Stree[2*MAX];
node?Min_Stree[2*MAX];
int?w[MAX];
int?getmax(int?a,int?b)
{
????return?a>b?a:b;????
}
int?getmin(int?a,int?b)
{
????return?a>b?b:a;????
}
int?Build_Max(int?now,int?l,int?r){
????Max_Stree[now].l=l;
????Max_Stree[now].r=r;
????if(l==r)Max_Stree[now].m=w[l];
????else?{
????????int?mid=(l+r)>>1;
????????int?max1=Build_Max(2*now,l,mid);
????????int?max2=Build_Max(2*now+1,mid+1,r);
????????Max_Stree[now].m=getmax(max1,max2);????
????}
????return?Max_Stree[now].m;????
}
int?Build_Min(int?now,int?l,int?r){
????Min_Stree[now].l=l;
????Min_Stree[now].r=r;
????if(l==r)Min_Stree[now].m=w[l];
????else?{
????????int?mid=(l+r)>>1;
????????int?min1=Build_Min(2*now,l,mid);
????????int?min2=Build_Min(2*now+1,mid+1,r);
????????Min_Stree[now].m=getmin(min1,min2);????
????}
????return?Min_Stree[now].m;????
}
int?Find_Max(int?now,int?l,int?r){
????int?left=Max_Stree[now].l;
????int?right=Max_Stree[now].r;
????if(left==l&&right==r)
????????return?Max_Stree[now].m;
????int?mid=(left+right)>>1;
????if(mid+1>r)return?Find_Max(2*now,l,r);
????if(mid<l)return?Find_Max(2*now+1,l,r);????
????else?return?getmax(Find_Max(2*now,l,mid),Find_Max(2*now+1,mid+1,r));
}
int?Find_Min(int?now,int?l,int?r){
????int?left=Min_Stree[now].l;
????int?right=Min_Stree[now].r;
????if(left==l&&right==r)
????????return?Min_Stree[now].m;
????int?mid=(left+right)>>1;
????if(mid+1>r)return?Find_Min(2*now,l,r);
????if(mid<l)return?Find_Min(2*now+1,l,r);????
????else?return?getmin(Find_Min(2*now,l,mid),Find_Min(2*now+1,mid+1,r));
}
posted on 2009-02-19 11:16
KNIGHT 閱讀(334)
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