UVa 10534 Wavio Sequence

题目链接 UVa 10534

分析：

求一个序列的最长先递增再递减的子序列，且递增和递减的长度相同

假设长度为 2 * n - 1，前 n 个递增，后 n 个递减，中间共享一个元素

LIS 要用 O(nlog2^n) 的做法，不然超时

LIS O(nlog2^n)做法

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 10000 + 10;
const int INF = 0X7fffffff;
int num[MAXN], dp1[MAXN], dp2[MAXN], g[MAXN];

int main()
{
int n;
while (~scanf("%d", &n))
{
for (int i = 1; i <= n; i++) scanf("%d", &num[i]);

memset(dp1, 0, sizeof(dp1));
memset(dp2, 0, sizeof(dp2));

for (int i = 0; i < MAXN; i++) g[i] = INF;
for (int i = 1; i <= n; i++) //正向求LIS
{
int k = lower_bound(g + 1, g + n + 1, num[i]) - g;
dp1[i] = max(dp1[i - 1], k);
g[k] = num[i];
}

for (int i = 0; i < MAXN; i++) g[i] = INF;
for (int i = n; i >= 1; i--) //逆向求LIS
{
int k = lower_bound(g + 1, g + n + 1, num[i]) - g;
dp2[i] = max(dp2[i + 1], k);
g[k] = num[i];
}
int Max = 0;
for (int i = 1; i <= n; i++)
Max = max(Max, min(dp1[i], dp2[i]));
cout << 2 * Max - 1 << endl;
}
return 0;
}