最长上升子序列

 

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 7686    Accepted Submission(s): 1800


Problem Description
We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an, is it almost sorted?
 

 

Input
The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,…,an.

1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with n>1000.
 

 

Output
For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).
 

 

Sample Input
3 3 2 1 7 3 3 2 1 5 3 1 4 1 5
 

 

Sample Output
YES YES NO
 
 
复制代码
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <map>
 9 #include <vector>
10 using namespace std;
11 #define  ll long long 
12 #define  N   100009
13 #define  M 1000000000
14 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
15 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
16 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
17 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
18 #define  mem(a,b)  memset(a,b,sizeof(a))
19 #define  ph  push_back
20 int t,n;
21 int a[N],dp[N];
22 int main()
23 {
24     scanf("%d",&t);
25     while(t--)
26     {
27         scanf("%d",&n);
28         gep(i,1,n) scanf("%d",&a[i]);
29         int l1=1,l2=1l;
30         dp[l1]=a[1];
31         gep(i,2,n){
32             if(a[i]>=dp[l1]){
33                 l1++;
34                 dp[l1]=a[i];
35             }
36             else{
37                 int pos=upper_bound(dp+1,dp+1+l1,a[i])-dp;
38                 dp[pos]=a[i];
39             }
40         }
41         dp[l2]=a[n];
42         gepp(i,n-1,1){
43             if(a[i]>=dp[l2]){
44                 l2++;
45                 dp[l2]=a[i];
46             }
47             else{
48                 int pos=upper_bound(dp+1,dp+1+l2,a[i])-dp;
49                 dp[pos]=a[i];
50             }
51         }
52         if(l1>=n-1||l2>=n-1){
53             printf("YES\n");
54         }
55         else{
56             printf("NO\n");
57         }
58     }
59     return 0;
60 }
复制代码

 

 

复制代码
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <map>
 9 #include <vector>
10 #include <stack>
11 using namespace std;
12 #define  ll long long 
13 #define  N   100009
14 #define  M 1000000000
15 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
16 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
17 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
18 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
19 #define  mem(a,b)  memset(a,b,sizeof(a))
20 #define  ph  push_back
21 int n,a[N],dp[N];
22 int pre[N];
23 stack<int>s;
24 vector<int>ve;
25 int main()
26 {
27     scanf("%d",&n);
28     gep(i,1,n) scanf("%d",&a[i]);
29     gep(i,1,n) dp[i]=1;
30     mem(pre,-1);
31     //递增
32     gep(i,2,n){
33         gep(j,1,i-1){
34             if(a[i]>a[j]&&dp[i]<dp[j]+1){
35                 dp[i]=dp[j]+1;//1~i
36                 pre[i]=j;
37             }
38         }
39     }
40     while(j!=-1){
41         s.push(a[j]);
42         j=pre[j];
43     }
44     int v=s.top();
45         s.pop();
46         printf("%d",v);
47     while(!s.empty()){
48         int v=s.top();
49         s.pop();
50         printf(" %d",v);
51     }
52     printf("\n");
53     /*
54     5
55     6 3 5 2 9
56     3 5 9
57     */
58     
59     //递减
60     gepp(i,n-1,1){
61         gep(j,i+1,n){
62             if(a[i]>a[j]&&dp[i]<dp[j]+1){
63                 dp[i]=dp[j]+1;//i~n
64                 pre[i]=j;
65             }
66         }
67     }
68     int MAX=0,j=1;
69     gep(i,1,n){
70         if(MAX<dp[i]){
71             MAX=dp[i];
72             j=i;
73         }
74     }
75     while(j!=-1){
76         ve.ph(a[j]);
77         j=pre[j];
78     }
79     gep(i,0,ve.size()-1){
80         printf("%d%c",ve[i],i==ve.size()-1?'\n':' ');
81     }
82     /*
83     5
84     5 1 3 2 6
85     5 3 2
86     */
87 
88     return 0;
89 }
复制代码

 

 

 

//ACM训练联盟周赛

  •  65536K
 

Teemo starts to do homework everyday. Today, he meets a hard problem when doing his homework.

There's an array A which contains n integers(for every 1<=i<=n, A[i] = 1 or A[i]= 2), you can choose an interval [l,r](1<=l<=r<=n), then reverse it so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

Input Format

  • The first line of the input contains an integer T(1=<T<=10), giving the number of test cases.
  • For every test case, the first line contains an integer n(1<=n<=2000). The second line contains n integers. The i th integer represents A[i](1<=A[i]<=2). 

Output Format

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

样例输入

1
4
1 2 1 2

样例输出

4


复制代码
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <map>
 9 #include <vector>
10 #include <stack>
11 using namespace std;
12 #define  ll long long 
13 #define  N   2009
14 #define  M 1000000000
15 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
16 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
17 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
18 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
19 #define  mem(a,b)  memset(a,b,sizeof(a))
20 #define  ph  push_back
21 int t,n,dpx[N][N],dpy[N][N];
22 int dp[N],a[N];
23 int l;
24 int main()
25 {
26     scanf("%d",&t);
27     
28     while(t--)
29     {
30         scanf("%d",&n);
31         gep(i,1,n) scanf("%d",&a[i]);
32         mem(dpx,0);mem(dpy,0);
33         gep(i,1,n){
34             l=0;
35             mem(dp,0);
36             gep(j,i,n){
37                 if(a[j]>=dp[l]){
38                     l++;    
39                     dp[l]=a[j];
40                 }
41                 else{
42                     int pos=upper_bound(dp+1,dp+1+l,a[j])-dp;
43                     dp[pos]=a[j];
44                 }
45                 dpx[i][j]=l;//i~j的最长上升子序列
46                 }
47         }
48         gepp(i,n,1){
49             l=0;
50             mem(dp,0);
51             gepp(j,i,1){
52                 if(a[j]>=dp[l]){
53                     l++;    
54                     dp[l]=a[j];
55                 }
56                 else{
57                     int pos=upper_bound(dp+1,dp+1+l,a[j])-dp;
58                     dp[pos]=a[j];
59                 }
60                 dpy[j][i]=l;//j~i的最长下降子序列
61             }
62         }
63         int MAX=0;
64         gep(i,1,n){
65             gep(j,1,n){
66                 MAX=max(MAX,dpx[1][n]-dpx[i][j]+dpy[i][j]);//reverse
67             }
68         }
69         printf("%d\n",MAX);
70     }
71     return 0;
72 }
复制代码

 

posted on   cltt  阅读(154)  评论(0)  编辑  收藏  举报

编辑推荐：
· go语言实现终端里的倒计时
· 如何编写易于单元测试的代码
· 10年+ .NET Coder 心语，封装的思维：从隐藏、稳定开始理解其本质意义
· .NET Core 中如何实现缓存的预热？
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
阅读排行：
· 分享一个免费、快速、无限量使用的满血 DeepSeek R1 模型，支持深度思考和联网搜索！
· 基于 Docker 搭建 FRP 内网穿透开源项目（很简单哒）
· ollama系列01：轻松3步本地部署deepseek，普通电脑可用
· 25岁的心里话
· 按钮权限的设计及实现
< 2025年3月 >
日 一 二 三 四 五 六
23 24 25 26 27 28 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 1 2 3 4 5

导航

统计

点击右上角即可分享
微信分享提示
SQL AI 助手