UVA

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;
}