[POJ1631]Bridging signals
题目大意:不知,根据样例猜测为最长上升子序列(竟然还对了)
题解:$O(n log_2 n)$,求二维偏序,(q为存答案的序列,a存原序列,len为答案)
for(int i = 1; i <= n; i++) { if(a[i] > q[len]) {q[++len]=a[i];continue;} *upper_bound(q, q + len, a[i]) = a[i]; }
卡点:无
C++ Code:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int T, n ,a; int f[40010], cnt; int main() { scanf("%d", &T); while (T--) { scanf("%d", &n); memset(f, 0x3f, sizeof f); for (int i = 0 ; i < n; i++) { scanf("%d", &a); *lower_bound(f, f + n, a) = a; } printf("%d\n",lower_bound(f, f + n, 0x3f3f3f3f) - f); } return 0; }