最长上升子序列

时间：2018-08-17 12:54:25 阅读：181 评论：0 收藏：0 [点我收藏+]

标签：first sam familiar rev selection panel des sorted msu

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

Input

The first line contains an integer

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

 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 }

 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 }

//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

 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 }

最长上升子序列

标签：first sam familiar rev selection panel des sorted msu

原文地址：https://www.cnblogs.com/tingtin/p/9492612.html

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