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?
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.
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
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 "); 54 } 55 else{ 56 printf("NO "); 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(" "); 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?' ':' '); 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
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
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 ",MAX); 70 } 71 return 0; 72 }