UVA 10534 Wavio Sequence (双向LIS)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475

思路：以a[i]为Wavio Sequence的最大值，（根据定义）计算从0到i和从i到n的lis长度的较小者，为以a[i]为最大值的Wavio Sequence的半长度，最后取所有半长度的最大值*2-1

完整代码：

/*0.102s*/#include<bits/stdc++.h>using namespace std;const int mx = 10005;int a[mx], d[2][mx], lis[mx];int main(){int N, len, i, j, ans;while (~scanf("%d", &N)){len = 0;for (i = 1; i <= N; ++i){scanf("%d", &a[i]);d[0][i] = j = lower_bound(lis + 1, lis + len + 1, a[i]) - lis;len = max(len, j);lis[j] = a[i];}len = 0;for (i = N; i; --i){d[1][i] = j = lower_bound(lis + 1, lis + len + 1, a[i]) - lis;len = max(len, j);lis[j] = a[i];}ans = 0;for (i = 1; i <= N; ++i)ans = max(ans, (min(d[0][i], d[1][i]) << 1) - 1);printf("%d\n", ans);}return 0;}