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            posts - 74,  comments - 33,  trackbacks - 0
            Sliding Window
            Time Limit: 12000MS Memory Limit: 65536K
            Total Submissions: 7213 Accepted: 1859
            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?-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 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 幾個月前的浙大月賽題的區間最大最小值,用隊列維護的方法還是不會
            用線段樹AC,有點郁悶,我的半吊子線段樹啊?? 居然10s才ac,不過幸好服務器沒掛。。。。。。
            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) 評論(0)  編輯 收藏 引用
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