poj 2823 Sliding Windows 线段树|单调队列

题意：

给你一个长为n的数组求任意连续长为k的子区域最大最小值

   区间极值，很裸的线段树。但是正解应该是单调队列，单调队列比线段树快近一倍，因为充分利用之前计算的值。

线段树差点超时

ACcode：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define maxn 1000100
#define tmp (st<<1)
#define mid ((l+r)>>1)
#define lson l,mid,tmp
#define rson mid+1,r,tmp|1
#define pushup1(x) maxx[x]=max(maxx[tmp],maxx[tmp|1])
#define pushup2(x) minn[x]=min(minn[tmp],minn[tmp|1])
using namespace std;
int maxx[maxn<<2],minn[maxn<<2];
inline void build(int l,int r,int st){
    if(l==r){
        scanf("%d",&maxx[st]);
        minn[st]=maxx[st];
        return ;
    }
    build(lson);
    build(rson);
    pushup1(st);
    pushup2(st);
}
inline int query(int L,int R,bool t,int l,int r,int st){
    if(L<=l&&r<=R)return t?maxx[st]:minn[st];
    if(t){
        int ret=-0x3f3f3f3f;
        if(L<=mid)ret=max(ret,query(L,R,t,lson));
        if(R>mid)ret=max(ret,query(L,R,t,rson));
        return ret;
    }
    else {
        int ret=0x3f3f3f3f;
        if(L<=mid)ret=min(ret,query(L,R,t,lson));
        if(R>mid)ret=min(ret,query(L,R,t,rson));
        return ret;
    }
}
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    build(1,n,1);
    for(int i=k;i<=n;++i)
        printf("%d%c",query(i-k+1,i,0,1,n,1),i==n?'\n':' ');
    for(int i=k;i<=n;++i)
        printf("%d%c",query(i-k+1,i,1,1,n,1),i==n?'\n':' ');
    return 0;
}

利用优先队列实现单调队列

ACcode：

#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#define maxn 1000100
using namespace std;
int a[maxn],maxx[maxn],minn[maxn],tot1,tot2;
struct cmp1{
    bool operator()(const int x,const int y){
        return a[x]>a[y];
    }
};
struct cmp2{
    bool operator()(const int x,const int y){
        return a[x]<a[y];
    }
};
priority_queue<int ,vector<int>,cmp1> dp1;
priority_queue<int ,vector<int>,cmp2> dp2;
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
        if(i<=k){
            dp1.push(i);
            dp2.push(i);
        }
    }
    tot1=tot2=0;
    maxx[++tot1]=a[dp1.top()];
    minn[++tot2]=a[dp2.top()];
    for(int i=k+1;i<=n;++i){
        dp1.push(i);
        dp2.push(i);
        while(i-dp1.top()>=k)dp1.pop();
        maxx[++tot1]=a[dp1.top()];
        while(i-dp2.top()>=k)dp2.pop();
        minn[++tot2]=a[dp2.top()];
    }
    for(int i=1;i<=n-k+1;++i)printf("%d%c",maxx[i],(i==n-k+1)?'\n':' ');
    for(int i=1;i<=n-k+1;++i)printf("%d%c",minn[i],(i==n-k+1)?'\n':' ');
    return 0;
}