n的演算法

通過1.彈出隊首 2.壓入隊尾 3.彈出隊尾 來保證區間內的單調性

P1886 滑動視窗 /【模板】單調佇列 - 洛谷 | 電腦科學教育新生態 (luogu.com.cn)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e6+10;
 4 struct node
 5 {
 6     int num,poi;
 7 }q[N];
 8 struct nod
 9 {
10     int num,poi;
11 }p[N];
12 int n,k;
13 int a[N];
14 int
 
 main()
15 {
16     cin>>n>>k;
17     for(int i=1;i<=n;++i)
18     cin>>a[i];
19     int head=1,tail=1;
20     q[head].num=a[1];
21     q[tail].poi=1;
22     for(int i=1;i<=n;++i)
23     {
24         //cout<<q[head].num<<endl;
25         while(head<=tail&&a[i]<q[tail].num)
 
26         tail--;
27         tail++;
28         q[tail]=(node){a[i],i};
29         while(head<=tail&&q[tail].poi-q[head].poi+1>k)
30         head++;
31         if(i>=k)
32         cout<<q[head].num<<" ";
33     }
34     cout<<endl;
35     head=1;tail=1;
36     p[head].num=a[1
 
];
37     p[head].poi=1;
38     for(int i=1;i<=n;++i)
39     {
40         while(head<=tail&&a[i]>p[tail].num)
41         tail--;
42         tail++;
43         p[tail]=(nod){a[i],i};
44         while(head<=tail&&p[tail].poi-p[head].poi+1>k)
45         head++;
46         if(i>=k)
47         cout<<p[head].num<<" ";
48     }
49     return 0;
50 }

P2216 [HAOI2007]理想的正方形 - 洛谷 | 電腦科學教育新生態 (luogu.com.cn)

 1 //雙重單調佇列
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 namespace _xzy
 5 {
 6     typedef long long ll;
 7     inline int read()
 8     {
 9         ll sm=0,flag=1;
10         char ch=getchar();
11         while(ch<'0'||ch>'9'){if(ch=='-')flag=-1;ch=getchar();}
12         while(ch>='0'&&ch<='9'){sm=sm*10+ch-'0';ch=getchar();}
13         return sm*flag;
14     }
15     ll a,b,n,ans=1e9;
16     ll g[1002][1002];
17     struct node
18     {
19         ll poi,num;
20     }e[1002],d[1002];
21     ll maxn[1002][1002],minn[1002][1002];
22     void work()
23     {
24         for(ll i=1;i<=a;++i)
25         {
26             ll head=1,tail=1,h=1,t=1;
27             e[head]=(node){1,g[i][1]};d[head]=(node){1,g[i][1]};
28             for(ll j=1;j<=b;++j)
29             {
30                 while(head<=tail&&e[tail].num<g[i][j])tail--;
31                 tail++;e[tail]=(node){j,g[i][j]};
32                 while(head<=tail&&e[tail].poi-e[head].poi+1>n)head++;
33                 
34                 while(h<=t&&d[t].num>g[i][j])t--;
35                 t++;d[t]=(node){j,g[i][j]};
36                 while(h<=t&&d[t].poi-d[h].poi+1>n)h++;
37                 
38                 if(j>=n)
39                 {
40                     maxn[i][j]=e[head].num;minn[i][j]=d[h].num;
41                 }
42             }
43         }
44         for(ll i=n;i<=a;++i)
45         for(ll j=n;j<=b;++j)
46         {
47             ll ma=0,mi=1e9;
48             for(ll k=i;k>=i-n+1;--k)
49             {
50                 ma=max(ma,maxn[k][j]);mi=min(mi,minn[k][j]);
51             }
52             ans=min(ans,abs(ma-mi));
53         }
54         cout<<ans;
55     }
56     void My_main()
57     {
58         a=read();b=read();n=read();
59         for(ll i=1;i<=a;++i)
60         for(ll j=1;j<=b;++j)
61         g[i][j]=read();
62         work();
63     }
64 }
65 int main()
66 {
67     _xzy::My_main();
68     return 0;
69 }