Uva--10534（动规，LIS，二分）
2014-08-02 19:32:43
Problem D
Wavio Sequence
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
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 Output for 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
|
9 9 1
|
思路：从头开始扫一遍LIS，再从尾开始扫一遍LIS。由于这里的数据范围比较大，应该采用二分优化的LIS算法，时间复杂度为O（nlogn），具体见程序。
1 /************************************************************************* 2 3 > File Name: j.cpp 4 > Author: Nature 5 > Mail: 564374850@qq.com 6 > Created Time: Thu 31 Jul 2014 12:21:05 PM CST 7 ************************************************************************/ 8 9 #include <cstdio> 10 #include <cstring> 11 #include <cstdlib> 12 #include <cmath> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 17 int B_search(int v,int *s,int len){ 18 int mid,l = 0,r = len; 19 while(l < r){ 20 mid = l + (r - l) / 2; 21 if(s[mid] == v) 22 return mid; 23 else if(s[mid] > v) 24 r = mid; 25 else 26 l = mid + 1; 27 } 28 return l; 29 } 30 31 int main(){ 32 //freopen("in","r",stdin); 33 int n; 34 int v[10005]; 35 int v2[10005]; 36 int dp1[10005]; 37 int dp2[10005]; 38 int len1[10005]; 39 int len2[10005]; 40 while(scanf("%d",&n) == 1){ 41 memset(dp1,0,sizeof(dp1)); 42 memset(dp2,0,sizeof(dp2)); 43 memset(len1,0,sizeof(len1)); 44 memset(len2,0,sizeof(len2)); 45 for(int i = 0; i < n; ++i){ 46 scanf("%d",&v[i]); 47 v2[n - i - 1] = v[i]; 48 } 49 for(int i = 0; i < n; ++i){ 50 if(i == 0){// || len1[i - 1] == 1){ 51 dp1[len1[i]++] = v[i]; 52 } 53 else if(dp1[len1[i - 1] - 1] < v[i]){ 54 dp1[len1[i - 1]] = v[i]; 55 len1[i] = len1[i - 1] + 1; 56 } 57 else{ 58 int pos = B_search(v[i],dp1,len1[i - 1]); 59 //if(dp1[pos] != v[i]); 60 dp1[pos] = v[i]; 61 len1[i] = len1[i - 1]; 62 } 63 } 64 for(int i = 0; i < n; ++i){ 65 if(i == 0){// || len2[i + 1] == 1){ 66 dp2[len2[i]++] = v2[i]; 67 } 68 else if(dp2[len2[i - 1] - 1] < v2[i]){ 69 dp2[len2[i - 1]] = v2[i]; 70 len2[i] = len2[i - 1] + 1; 71 } 72 else{ 73 int pos = B_search(v2[i],dp2,len2[i - 1]); 74 //if(dp2[pos] != v2[i]) 75 dp2[pos] = v2[i]; 76 len2[i] = len2[i - 1]; 77 } 78 } 79 int tmax = 1; 80 for(int i = 0; i < n; ++i){ 81 if(len1[i] > 1 && len2[n - i - 1] > 1 && len1[i] == len2[n - i - 1] && (len1[i] + len2[n - i - 1] - 1) % 2) 82 tmax = max(tmax,len1[i] + len2[n - i - 1] - 1); 83 } 84 printf("%d\n",tmax); 85 } 86 return 0; 87 }
