标签:ems 去掉 owb eve 分析 fine sign print mil
题意:对于一个序列,要求去掉正好K个数字,若能使其成为不上升子序列或不下降子序列,则“A is a magic array.”,否则"A is not a magic array.\n"。
分析:
1、求一遍LCS,然后在将序列逆转,求一遍LCS,分别可得最长上升子序列和最长下降子序列的长度tmp1、tmp2。
2、n - tmp1 <= k或n - tmp2 <= k即可,需要去掉的去完之后,在已经是最长上升或最长下降的序列中随便去够k个就好了。
#include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define lowbit(x) (x & (-x)) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 100000 + 10; const int MAXT = 10000 + 10; using namespace std; int a[MAXN], dp[MAXN]; int main(){ int T; scanf("%d", &T); while(T--){ int n, k; scanf("%d%d", &n, &k); for(int i = 0; i < n; ++i){ scanf("%d", &a[i]); } memset(dp, INT_INF, sizeof dp); for(int i = 0; i < n; ++i){ *lower_bound(dp, dp + n, a[i]) = a[i]; } int tmp1 = lower_bound(dp, dp + n, INT_INF) - dp; if(n - tmp1 <= k){ printf("A is a magic array.\n"); } else{ reverse(a, a + n); memset(dp, INT_INF, sizeof dp); for(int i = 0; i < n; ++i){ *lower_bound(dp, dp + n, a[i]) = a[i]; } int tmp2 = lower_bound(dp, dp + n, INT_INF) - dp; if(n - tmp2 <= k){ printf("A is a magic array.\n"); } else{ printf("A is not a magic array.\n"); } } } return 0; }
HDU - 6197 array array array (最长上升子序列&最长下降子序列)
标签:ems 去掉 owb eve 分析 fine sign print mil
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/7517566.html