poj 2823 Sliding Window

题目链接如下：

2823 -- Sliding Window

目录

1.线段树

2.单调栈

---

1.线段树

单单用一个维护区间最大值最小值的线段树就可以方便地解决这个问题。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<vector>
#include<algorithm>
#include<cstring>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cmath>
#include<climits>

using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
typedef long long ll;
const int MAXN=1000001;
int max_tree[MAXN<<2],min_tree[MAXN<<2];
int max_ans[MAXN],min_ans[MAXN];

void push_up(int rt){
    max_tree[rt]= max(max_tree[rt<<1],max_tree[rt<<1|1]);
    min_tree[rt]= min(min_tree[rt<<1],min_tree[rt<<1|1]);
}

void build(int l,int r,int rt){
    if (l==r){
        scanf("%d",&max_tree[rt]);
        min_tree[rt]=max_tree[rt];
        return;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(rt);
}

void query(int pos,int a,int b,int l,int r,int rt){
    if (a<=l && r<=b){
        max_ans[pos]= max(max_ans[pos],max_tree[rt]);
        min_ans[pos]= min(min_ans[pos],min_tree[rt]);
        return;
    }
    int mid=(l+r)>>1;
    if (a<=mid){
        query(pos,a,b,lson);
    }
    if (b>mid){
        query(pos,a,b,rson);
    }
}

int main(){
    int n,k;
    while (~scanf("%d %d",&n,&k)){
        build(1,n,1);
        for (int i = 0; i <= n; ++i) {
            min_ans[i]=INT_MAX;
            max_ans[i]=INT_MIN;
        }
        int pos=0;
        for (int i = 1; i <= n-k+1; ++i) {
            query(pos++,i,i+k-1,1,n,1);
        }
        for (int i = 0; i < pos; ++i) {
            printf("%d ",min_ans[i]);
        }
        printf("\n");
        for (int i = 0; i < pos; ++i) {
            printf("%d ",max_ans[i]);
        }
        printf("\n");
    }
    return 0;
}

2.单调栈

除了线段树，这题也能用单调栈解决。

用一个双向队列维护区间内的最大值最小值。front的位置用来维护区间的大小，并且维护的区间中的front就是最大最小值，因为没被弹出，或者说是新压入的值，back的值用来维护这个单调栈，最大的值我们维护递减序列，因此在压入的时候把小于等于的弹出，相反也是一样的思路。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<vector>
#include<algorithm>
#include<cstring>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cmath>
#include<climits>

using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
typedef long long ll;
const int MAXN=1000001;
struct node{
    int pos;
    int val;
}t;
deque<node> max_q,min_q;
int max_ans[MAXN],min_ans[MAXN];
int main(){
    int n,k,x;
    while (~scanf("%d %d",&n,&k)){
        max_q.clear();
        min_q.clear();
        for (int i = 1; i < k; ++i) {
            scanf("%d",&x);
            while (!max_q.empty() && max_q.back().val<=x) max_q.pop_back();
            while (!min_q.empty() && min_q.back().val>=x) min_q.pop_back();
            t.pos=i,t.val=x;
            max_q.push_back(t);
            min_q.push_back(t);
        }
        for (int i = k; i <= n; ++i) {
            while (!max_q.empty() && max_q.front().pos<i-k+1) max_q.pop_front();
            while (!min_q.empty() && min_q.front().pos<i-k+1) min_q.pop_front();
            scanf("%d",&x);
            while (!max_q.empty() && max_q.back().val<=x) max_q.pop_back();
            while (!min_q.empty() && min_q.back().val>=x) min_q.pop_back();
            t.pos=i,t.val=x;
            max_q.push_back(t);
            min_q.push_back(t);
            max_ans[i]=max_q.front().val;
            min_ans[i]=min_q.front().val;
        }
        for (int i = k; i <= n; ++i) {
            printf("%d ",min_ans[i]);
        }
        printf("\n");
        for (int i = k; i <= n; ++i) {
            printf("%d ",max_ans[i]);
        }
        printf("\n");
    }
    return 0;
}