题意:有两个长度分别为p+1和q+1的序列,每个序列中的各个元素互不相同,且都是1~n2的整数。两个序列的第一个元素均为1。求出A和B的最长公共子序列长度。
分析:
A = {1,7,5,4,8,3,9},B = {1,4,3,5,6,2,8,9}。
1、A中元素各不相同,因此将A中序列重新编号为1~p+1,即A中每个元素在A中是第几个出现的。
2、按照A中制定的编号原则给B重新编号,则B为{1,4,6,3,0,0,5,7},0表示这些元素在A中没有出现过。
3、新的A和B的LCS实际上就是新的B的LIS。LIS可在O(nlogn)内解决。
#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 = 62500 + 10; const int MAXT = 100000 + 10; using namespace std; int a[MAXN], b[MAXN], dp[MAXN]; map<int, int> mp; int main(){ int T; scanf("%d", &T); int kase = 0; while(T--){ mp.clear(); int n, p, q; scanf("%d%d%d", &n, &p, &q); for(int i = 1; i <= p + 1; ++i){ scanf("%d", &a[i]); mp[a[i]] = i; } for(int i = 1; i <= q + 1; ++i){ scanf("%d", &b[i]); if(!mp.count(b[i])){ b[i] = 0; } else{ b[i] = mp[b[i]]; } } memset(dp, INT_INF, sizeof dp); for(int i = 1; i <= q + 1; ++i){ *lower_bound(dp, dp + q + 1, b[i]) = b[i]; } printf("Case %d: %d ", ++kase, lower_bound(dp, dp + q + 1, INT_INF) - dp); } return 0; }