Sliding Window
Time Limit: 12000MS Memory Limit: 65536K
Total Submissions: 46075 Accepted: 13317
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 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

給一個長度為n的陣列，有一個長度為k的滑動視窗，滑動視窗從最左端開始向右移動，每個視窗中有最大值和最小值，求每次視窗滑動時，出現在視窗中的最小值和最大值。

這道題目也可以用線段樹做，不過時間卡的比較緊。。裸的最大最小值模板，套上去就可以用了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int
 
 N=1000010;
const int INF=0x3f3f3f3f;
int minx,maxx;

struct Tree{
    int l,r;
    int minx,maxx;
}tree[N<<2];

void build(int L,int R,int rt){
    tree[rt].l=L;
    tree[rt].r=R;
    if(tree[rt].l==tree[rt].r){
        scanf("%d",&tree[rt].minx);
        tree[rt].maxx=tree[rt].minx;
        return ;
    }
    int mid=(L+R)>>1;
    build(L,mid,rt*2);
    build(mid+1,R,rt*2+1);
    tree[rt].minx=min(tree[rt*2].minx,tree[rt*2+1].minx);
    tree[rt].maxx=max(tree[rt*2].maxx,tree[rt*2+1].maxx);
}


void QueMin(int L,int R,int rt){
    if(L<=tree[rt].l && tree[rt].r<=R){
        minx=min(minx,tree[rt].minx);
        return ;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(R<=mid)
        QueMin(L,R,rt*2);
    else if(L>=mid+1)
        QueMin(L,R,rt*2+1);
    else{
        QueMin(L,mid,rt*2);
        QueMin(mid+1,R,rt*2+1);
    }
}

void QueMax(int L,int R,int rt){
    if(L<=tree[rt].l && tree[rt].r<=R){
        maxx=max(maxx,tree[rt].maxx);
        return ;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(R<=mid)
        QueMax(L,R,rt*2);
    else if(L>=mid+1)
        QueMax(L,R,rt*2+1);
    else{
        QueMax(L,mid,rt*2);
        QueMax(mid+1,R,rt*2+1);
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    int n,k;
    while(~scanf("%d%d",&n,&k)){
        build(1,n,1);  //對1~n的範圍建樹
        int m=n-k,r;  //每次迴圈視窗的初始值到m就可以停止了
        for(int l=1;l<=m;l++){
            r=l+k-1;   // r是當前視窗最右邊的長度
            minx=INF;  //記得初始化
            QueMin(l,r,1); //訪問當前視窗的最小值
            printf("%d ",minx);
        }
        //最後一組的資料
        minx=INF;
        QueMin(n-k+1,n,1);
        printf("%d\n",minx);

        //同上，這次是處理最大值
        for(int l=1;l<=m;l++){
            r=l+k-1;
            maxx=-INF;
            QueMax(l,r,1);
            printf("%d ",maxx);
        }
        maxx=-INF;
        QueMax(n-k+1,n,1);
        printf("%d\n",maxx);

    }

    return 0;
}