单调队列 poj2823

题目链接：http://poj.org/problem?id=2823

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

题意：

给定数字n，和k     范围在区间【1，le6】    

下一行有n个数， 

依次输出每一个区间长度为k的           最小值

　　另一行依次输出区间长度为k的    最大值

 

 

 

解：　

　　显然这里用的是优先队列，又名单调队列。

　　顾名思义就是单调递增，或者是单调递减的队列。

 

　　　　至于单调队列最入门的用法，自然是求数组中的最小值或者最大值，这里求最大值为例。

#include<queue>
int main()
{
    int a[15];
    for (int i=1;i<=10;i++){
        a[i]=rand()%100+1;
    }
    for (int i=1;i<=10;i++){
        printf("a[%d]=%d
",i,a[i]);
    }
    queue<int>q;
    while (!q.empty())
        q.pop();
    q.push(a[1]);
    for (int i=2;i<=10;i++){
        if (q.front()<a[i]){
            q.pop();
            q.push(a[i]);
        }
    }
    cout<<"最大值为："<<q.front()<<endl;
}

 

 

回到正题，对于上题。

AC代码：   （神奇的地方：用G++提交会超时，用C++提交能AC）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<map> 
#include<algorithm>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double pi=acos(-1.0);

const int MAX=1e6+10;
struct s{
    int val,id;
};
s minq[MAX],maxq[MAX];
int maxhead,maxtail,minhead,mintail;
int maxans[MAX],minans[MAX];
int cnt;
 
int main()
{
    int n,k,tmp;
    scanf("%d%d",&n,&k);//    cin>>n>>k;
    cnt=0;
    maxhead=maxtail=minhead=maxtail=0;
    for(int i=0;i<k;i++){
        scanf("%d",&tmp);//cin>>tmp;
        while(maxhead<maxtail&&maxq[maxtail-1].val<=tmp)//递减队列 
            maxtail--;
        maxq[maxtail].val=tmp;
        maxq[maxtail++].id=i;
        
        while(minhead<mintail&&minq[mintail-1].val>=tmp)//递增队列
            mintail--;
        minq[mintail].val=tmp;
        minq[mintail++].id=i;    
    }
    
    for (int i=k;i<n;i++){
        maxans[cnt]=maxq[maxhead].val;
        minans[cnt++]=minq[minhead].val;
        scanf("%d",&tmp);//cin>>tmp;
        while (minhead<mintail&&i-minq[minhead].id>=k)
            minhead++;
        while (minhead<mintail&&minq[mintail-1].val>=tmp)
            mintail--;
        minq[mintail].val=tmp;
        minq[mintail].id=i;
        mintail++;

        while (maxhead < maxtail && i - maxq[maxhead].id >= k) ++maxhead;
        while (maxhead < maxtail && maxq[maxtail - 1].val <= tmp) --maxtail;
        maxq[maxtail].val = tmp;
        maxq[maxtail].id = i;
        ++maxtail;        
    }
    minans[cnt] = minq[minhead].val;
    maxans[cnt] = maxq[maxhead].val;
    ++cnt;
    for (int i=0;i<cnt;i++){
        if (i==0)
            printf("%d",minans[i]);
        else
            printf(" %d",minans[i]);
    }
    cout<<endl;
    for (int i=0;i<cnt;i++){
        if (i==0)
            printf("%d",maxans[i]);
        else
            printf(" %d",maxans[i]);
    }
    cout<<endl;
    return 0;
}