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BNUOJ 14381 Wavio Sequence
Wavio Sequence
64-bit integer IO format: %lld Java class name: Main
Wavio is a sequence of integers. It has some interesting properties.
Wavio is of odd length i.e. L = 2*n + 1.
The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.
Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be 9.
Input
The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.
Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will be N integers.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input
10
1 2 3 4 5 4 3 2 1 10
19
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1
5
1 2 3 4 5
Sample Output
9
9
1
解题：最长上升子序列加强版。单调队列优化！！！重点。先顺着求最长上升子序列，再逆着求最长上升子序列。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 10100;18 int dp1[maxn],dp2[maxn],d[maxn],q[maxn];19 int bsearch(int lt,int rt,int val) {20 while(lt <= rt) {21 int mid = (lt+rt)>>1;22 if(q[mid] < val) lt = mid+1;//严格上升单调取少于符号，上升的取少于等于23 else rt = mid-1;24 }25 return lt;26 }27 int main() {28 int n,i,j,head,tail;29 while(~scanf("%d",&n)) {30 for(i = 0; i < n; i++)31 scanf("%d",d+i);32 head = tail = 0;33 for(i = 0; i < n; i++) {34 if(head == tail) {35 q[head++] = d[i];36 dp1[i] = head-tail;37 }else if(d[i] > q[head-1]){38 q[head++] = d[i];39 dp1[i] = head-tail;40 }else{41 int it = bsearch(tail,head-1,d[i]);42 dp1[i] = it - tail + 1;43 q[it] = d[i];44 }45 }46 head = tail = 0;47 for(i = n-1; i >= 0; i--) {48 if(head == tail) {49 q[head++] = d[i];50 dp2[i] = head-tail;51 }else if(d[i] > q[head-1]){52 q[head++] = d[i];53 dp2[i] = head-tail;54 }else{55 int it = bsearch(tail,head-1,d[i]);56 dp2[i] = it - tail + 1;57 q[it] = d[i];58 }59 }60 int ans = 1;61 for(i = 0; i < n; i++){62 ans = max(ans,min(dp1[i],dp2[i])*2-1);63 }64 printf("%d\n",ans);65 }66 return 0;67 }