uva 10534 Wavio Sequence
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 length9. 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 be9.
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 beN integers.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input Output for Sample Input
101 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 59 9 1
题目大意：找出最长的回文子序列。
解题思路：先从头到尾找一遍最长上升子序列，，然后再从尾到头找一遍最长上升子序列，最后找出一点使得两者中最小的子序列长度最大，答案就是ans * 2 – 1。因为数据量比较大，所以dp最长上升子序列的时候不能用常规方法，要用栈和贪心的思想。（具体见代码）
#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#define N 10005using namespace std;typedef long long ll;int num[N], dp1[N], dp2[N], S1[N], S2[N];int main() {int n;while (scanf("%d", &n) != EOF) {for (int i = 0; i < n; i++) {scanf("%d", &num[i]);}int Max1 = 0, cnt1 = 0;for (int i = 0; i < n; i++) {if (cnt1 == 0) {S1[cnt1++] = num[i];dp1[0] = 1;continue;}if (S1[cnt1 – 1] < num[i]) {S1[cnt1++] = num[i];} else {int pos = lower_bound(S1, S1 + cnt1, num[i]) – S1;S1[pos] = num[i];}dp1[i] = cnt1;}int Max2 = 0, cnt2 = 0;for (int i = n – 1; i >= 0; i–) {if (cnt2 == 0) {S2[cnt2++] = num[i];dp2[n – 1] = 1;continue;}if (S2[cnt2 – 1] < num[i]) {S2[cnt2++] = num[i];} else {int pos = lower_bound(S2, S2 + cnt2, num[i]) – S2;S2[pos] = num[i];}dp2[i] = cnt2;}int temp = 999999999, ans;for (int i = 0; i < n; i++) {if (temp > abs(dp1[i] – dp2[i])) {temp = abs(dp1[i] – dp2[i]);ans = min(dp1[i], dp2[i]);}}printf("%d\n", ans * 2 – 1);}return 0;}
松树亭亭玉立的耸立在周围小草小花的中间，
