Sliding Window POJ

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

题意：找出每组连续的k个数的最大值最小值。

可以用线段树，也可以优先队列。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXN=1000000+10;
const int inf=0x3f3f3f;
int n,m;
struct node1
{
    ll x;
    int y;
    friend bool operator < (node1 a,node1 b)
    {
        return a.x>b.x;
    }
}z1;
struct node2
{
    ll x;
    int y;
    friend bool operator<(node2 a,node2 b)
    {
        return a.x<b.x;
    }
}z2;
priority_queue<node1>q1;
priority_queue<node2>q2;
ll z[MAXN];
ll minn[MAXN];
ll maxn[MAXN];
int main()
{
        scanf("%d%d",&n,&m);
        while(!q1.empty()) q1.pop();
        while(!q2.empty()) q2.pop();
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&z[i]);
            z1.y=z2.y=i;
            z2.x=z1.x=z[i];
            if(i<=m)
            {
                q1.push(z1);
                q2.push(z2);
            }
        }
        int ans=0;
        minn[ans]=q1.top().x;
        maxn[ans++]=q2.top().x;
        int left=2;
        for(int i=m+1;i<=n;i++)
        {
            z1.x=z[i];z1.y=i;
            z2.x=z[i];z2.y=i;
            q1.push(z1);
            q2.push(z2);
            while(q1.top().y<left)
            {
                q1.pop();
            }
            while(q2.top().y<left)
            {
                q2.pop();
            }
            left++;
            minn[ans]=q1.top().x;
            maxn[ans++]=q2.top().x;
        }
        printf("%lld",minn[0]);
        for(int i=1;i<ans;i++) printf(" %lld",minn[i]);
        printf("
");
        printf("%lld",maxn[0]);
        for(int i=1;i<ans;i++) printf(" %lld",maxn[i]);
        printf("
");
}

线段树

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define maxn 1000009
#define root  1,n,1
#define lson  l,mid,rt<<1
#define rson  mid+1,r,rt<<1|1
using namespace std;
struct node{
    int l,r,rt;
    int mid(){
        return (l+r)/2;
    }
}tree[maxn*4];
int n,k;
int sum1[maxn<<2],sum2[maxn<<2];
void pushup(int rt){
    sum1[rt]=min(sum1[rt<<1],sum1[rt<<1|1]);
    sum2[rt]=max(sum2[rt<<1],sum2[rt<<1|1]);
}
void build(int l,int r,int rt){
    tree[rt].l=l;tree[rt].r=r;
   if(l==r){
    tree[rt].l=tree[rt].r=l;
    scanf("%d",&sum1[rt]);
    sum2[rt]=sum1[rt];return ;
   }
   int mid=tree[rt].mid();
   build(lson);
   build(rson);
   pushup(rt);
}
int query1(int a,int b,int l,int r,int rt){
    if(a==l&&b==r){
        return sum1[rt];
    }
    int mid=tree[rt].mid();
    if(a>mid){
        return query1(a,b,mid+1,r,rt*2+1);
    }
    else if(b<=mid){
        return query1(a,b,l,mid,rt*2);
    }
    else
        return min(query1(a,mid,lson),query1(mid+1,b,rson));

}
int query2(int a,int b,int l,int r,int rt){
    if(a==l&&b==r){
        return sum2[rt];
    }
    int mid=tree[rt].mid();
    if(a>mid){
        return query2(a,b,mid+1,r,rt*2+1);
    }
    else if(b<=mid){
        return query2(a,b,l,mid,rt*2);
    }
    else
        return max(query2(a,mid,lson),query2(mid+1,b,rson));

}
int main(){
    mem(sum1,inf);mem(sum2,-inf);
    scanf("%d%d",&n,&k);
    build(root);
    for(int i=1;i<=n-k+1;i++){
            if(i!=1)printf(" ");
      printf("%d",query1(i,i+k-1,root));
    }
    printf("
");
    for(int i=1;i<=n-k+1;i++){
            if(i!=1)printf(" ");
      printf("%d",query2(i,i+k-1,root));
    }
    printf("
");
}