UVA-10534 Wavio Sequence (DP+LIS)

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思路：

正反各求出每个数作为结尾数字的LIS，然后找出最长的Wavio Sequence，用O(n²)试了一次TLE，所以只能用O(nlogn)的算法。

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Code:

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define M(a) memset(a, 0, sizeof(a))
const int maxn = 10000 + 10;
int bd[maxn], ed[maxn], bg[maxn], eg[maxn], a[maxn], b[maxn];

int main() {
    int n;
    while(~scanf("%d", &n)) {
        M(bd), M(ed), M(eg), M(bg);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            b[n-i+1] = a[i];
        }
        for (int i = 1; i <= n; ++i) bg[i] = eg[i] = INF;
        for (int i = 1; i <= n; ++i) {
            int k = lower_bound(bg+1, bg+n+1, a[i]) - bg;
            bd[i] = k;
            bg[k] = a[i];
            k = lower_bound(eg+1, eg+n+1, b[i]) - eg;
            ed[i] = k;
            eg[k] = b[i];
        }
        int ans = 0, temp;
        for (int i = 1; i <= n; ++i) {
            temp = min(bd[i], ed[n-i+1]);
            ans = max(ans, temp);
        }
        printf("%d\n", 2*ans-1);
    }

    return 0;
}