优先队列
指使用STL库的priority_queue进行模拟，优点在于实现简单。可用于求区间最值，由于使用堆操作，时间复杂度在([nlog_2{n}~n^2log_2{n}])，当数据较大时容易TLE
单调队列
单调队列使用STL的deque进行模拟，也可以用数组和双指针((head,tail))，有两种操作，删头和去尾，实现一个区间内单调增或减的队列，线性复杂度。单调队列经常与dp一起优化。
删头
判断头元素是否过期——超过窗口长度，过期即弹出；
去尾
判断尾元素是否大于新元素（递减）,小于新元素（递增）。原理是若尾元素若不满足条件，则尾元素一定不可能移动到头元素，因此可以弹出；
dp优化
对一定范围内的dp数组进行单调队列维护，目的在于取一定范围内的dp最值，而不仅仅是前一个
转移方程一般为 (dp[i]=dp[j]+w[i],(j+m<i))
经典题：
洛谷P1440 求m区间内的最小值
解法一：
经典ST表，MLE。需要的二维数组范围太大，超过内存限制，速度也比较慢
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define MAXN 2000020
#define MOD 10007
#define sf(a,b) read(a),read(b)
using namespace std;
int n,m,brr[MAXN][(int)log2(MAXN)+1];
int main()
{//MLE
scanf("%d%d",&n,&m);
FOR2(i,1,n)
{
scanf("%d",&brr[i][0]);
}
int logm=log2(m);
for(int h=1;h<=logm;h++)
{
for(int i=1;i+(1<<h)-1<=n;i++)
{
brr[i][h]=min(brr[i][h-1],brr[i+(1<<h-1)][h-1]);
}
}
int minn=INF;
cout<<0<<endl;
for(int i=2;i<=n;i++)
{
if(i<=m+1){
minn=min(minn,brr[i-1][0]);
cout<<minn<<endl;
continue;
}
cout<<min(brr[i-m][logm],brr[i-(1<<logm)][logm])<<endl;
}
return 0;
}
解法二 ：
优化内存ST表，TLE，使用一层数组重复计算值，解决了内存的问题，但是ST表迭代计算速度慢的劣势导致TLE
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define MAXN 2000020
#define MOD 10007
#define sf(a,b) read(a),read(b)
using namespace std;
int arr[MAXN],n,m,brr[MAXN];
int main()
{//TLE
scanf("%d%d",&n,&m);
FOR2(i,1,n)
{
scanf("%d",&brr[i]);
arr[i]=brr[i];
}
int logm=log2(m);
for(int h=1;h<=logm;h++)
{
for(int i=1;i+(1<<h)-1<=n;i++)
{
brr[i]=min(brr[i],brr[i+(1<<h-1)]);
}
}
int minn=INF;
printf("0
");
for(int i=2;i<=n;i++)
{
if(i<=m+1){
minn=min(minn,arr[i-1]);
printf("%d
",minn);
continue;
}
printf("%d
",min(brr[i-m],brr[i-(1<<logm)]));
}
return 0;
}
解法三：
优先队列，550ms过，但是根堆的调整速度还是略慢
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define MAXN 2000020
#define MOD 10007
#define sf(a,b) read(a),read(b)
using namespace std;
typedef struct{
int val;
int pos;
}NODE;NODE nodes[MAXN];
int n,m;
struct cmp{
bool operator() (NODE n1,NODE n2)
{
return n1.val>=n2.val; //小根堆
}
};
priority_queue<NODE,vector<NODE>,cmp>q;
int main()
{
scanf("%d%d",&n,&m);
int minp=0,minn=INF;
FOR2(i,1,n)
{
scanf("%d",&nodes[i].val);
nodes[i].pos=i;
if(i==1)cout<<0<<endl;
else
{
while(m+q.top().pos<i)q.pop();//是否过期
printf("%d
",q.top().val);
}
if(i!=n)q.push(nodes[i]);
}
return 0;
}
解法四：
单调队列，400ms,线性复杂度，为正解
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define freopenin(a) freopen(a,"r",stdin);
#define freopenout(a) freopen(a,"w",stdout);
#define MAXN 2000100
#define MOD 10007
using namespace std;
typedef struct{
int val,pos;
}NODE;NODE nodes[MAXN];
int n,m;
deque<NODE>q;//递增
int main()
{
scanf("%d%d",&n,&m);
FOR2(i,1,n)
{
scanf("%d",&nodes[i].val);
nodes[i].pos=i;
while(!q.empty()&&q.back().val>=nodes[i].val)q.pop_back();
while(!q.empty()&&q.front().pos+m<=i)q.pop_front();
q.push_back(nodes[i]);
if(i==1) printf("0
");
if(i!=n)
{
printf("%d
",q.front().val);
}
}
return 0;
}
洛谷P1886 滑动窗口
解法一：
优先队列，根堆调整复杂度(O(nlog_2{n}))的劣势导致根本过不了第二个点，直接TLE
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define freopenin(a) freopen(a,"r",stdin);
#define freopenout(a) freopen(a,"w",stdout);
#define MAXN 1000100
#define MOD 10007
using namespace std;
typedef struct{
int val,pos;
}NODE;NODE nodes[MAXN];
struct cmp{
bool operator() (NODE n1,NODE n2)
{
return n1.val>=n2.val;
}
};
struct cmp2{
bool operator() (NODE n1,NODE n2)
{
return n1.val<=n2.val;
}
};
priority_queue<NODE,vector<NODE>,cmp>q;//小根堆
priority_queue<NODE,vector<NODE>,cmp2>q2;//大根堆
int ansx[MAXN];
int ansn[MAXN];
int n,k;
int main()
{
cin>>n>>k;
FOR2(i,1,n)
{
scanf("%d",&nodes[i].val);
nodes[i].pos=i;
q.push(nodes[i]);
q2.push(nodes[i]);
if(i-k>=0)
{
while(q.top().pos+k<=i)q.pop();
while(q2.top().pos+k<=i)q2.pop();
ansx[i-k]=q2.top().val;
ansn[i-k]=q.top().val;
}
}
FOR(i,0,n-k+1)printf("%d ",ansn[i]);
printf("
");
FOR(i,0,n-k+1)printf("%d ",ansx[i]);
return 0;
}
解法二：
单调队列，550ms过，线性复杂度NB
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x3f3f3f3f;
#define freopenin(a) freopen(a,"r",stdin);
#define freopenout(a) freopen(a,"w",stdout);
#define MAXN 1000100
#define MOD 10007
using namespace std;
typedef struct{
int val,pos;
}NODE;NODE nodes[MAXN];
deque<NODE>q1;//递减
deque<NODE>q2;//递增
int ansx[MAXN],ansn[MAXN];
int n,m;
int main()
{
scanf("%d%d",&n,&m);
FOR(i,0,n)
{
scanf("%d",&nodes[i].val);
nodes[i].pos=i;
while(!q1.empty()&&q1.back().val>=nodes[i].val)q1.pop_back();
while(!q2.empty()&&q2.back().val<=nodes[i].val)q2.pop_back();
while(!q1.empty()&&q1.front().pos+m<=i)q1.pop_front();
while(!q2.empty()&&q2.front().pos+m<=i)q2.pop_front();
q1.push_back(nodes[i]);
q2.push_back(nodes[i]);
ansx[i]=q1.front().val;
// for(int i=0;i<q1.size();i++)cout<<q1[i].val<<" ";
// cout<<q1.size()<<endl;
// cout<<endl;
ansn[i]=q2.front().val;
}
FOR(i,m-1,n)printf("%d ",ansx[i]);
printf("
");
FOR(i,m-1,n)printf("%d ",ansn[i]);
return 0;
}
JZOJ P2944烽火传递
题意
一串数字，每个窗口m内必须至少有两个点被选中，问选中点的和最小值
基本思路
使用单调队列对dp数组优化，使得当前选择的点加上之前m个状态的最小值，dp求和。
转移方程 (dp[i]=dp[j]+w[i], (i-m<j<i))
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f;
#define freopenin(a) freopen(a,"r",stdin);
#define freopenout(a) freopen(a,"w",stdout);
#define MAXN 1000100
#define MOD 10007
using namespace std;
typedef struct{
ll val,pos;
}NODE;NODE dp[MAXN];
ll w[MAXN],n,m;
deque<NODE>q;
int main()
{
scanf("%d%d",&n,&m);
dp[0]={0,0};q.push_back(dp[0]);
FOR2(i,1,n)
{
scanf("%d",&w[i]);
dp[i].pos=i;
while(!q.empty()&&q.front().pos+m<i) q.pop_front();
dp[i].val=q.front().val+w[i];
while(!q.empty()&&q.back().val>dp[i].val) q.pop_back();
q.push_back(dp[i]);
}
ll ans=INF;
FOR2(i,n-m+1,n)
{
ans=min(ans,dp[i].val);
}
printf("%d
",ans);
return 0;
}
附使用数组和指针的手动单调队列，速度快20ms
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<bits/stdc++.h>
using namespace std;
int n,m;
int w[100001];
int que[100001],head=0,tail=0;
int f[100001];
int main()
{
scanf("%d%d",&n,&m);
int i,j;
for (i=1;i<=n;++i)
scanf("%d",&w[i]);
memset(f,127,sizeof f);
f[0]=0;
que[0]=0;
for (i=1;i<=n;++i)
{//对f做单调队列
if (que[head]<i-m)
++head;//将超出范围的队头删掉
f[i]=f[que[head]]+w[i];//转移（用队头）
while (head<=tail && f[que[tail]]>f[i])
--tail;//将不比它优的全部删掉
que[++tail]=i;//将它加进队尾
}
for(int i=0;i<=n;i++)
{
cout<<que[i]<<" ";
}
cout<<endl;
for(int i=0;i<=n;i++)
{
cout<<f[i]<<" ";
}
cout<<endl;
int ans=0x7f7f7f7f;
for (i=n-m+1;i<=n;++i)
ans=min(ans,f[i]);
printf("%d
",ans);
}
其他题目：
洛谷：P3957 跳房子
P2216 [HAOI2007]理想的正方形
P1725琪露诺
P1714切蛋糕
【Tyvj1305】最大最大子序和
【Vijos1243】生产产品
【Hdu3530】Subsequence
【Hdu3401】Subsequence
【Poj1742】Coins
[【Hdu4374】 One hundred layer]
【CodeForces372C】 Watching Fireworks is Fun
参考博客
https://blog.csdn.net/hjf1201/article/details/78729320
https://www.cnblogs.com/Dxy0310/p/9742045.html
https://sweetlemon.blog.luogu.org/dan-diao-dui-lie
https://blog.csdn.net/weixin_44412226/article/details/90476669