Prince and Princess UVA

题目传送门

题意：求两个序列的最长公共子序列。

思路：如果用最朴素的算法，显然会超时，所以用LIS的方式对LCS进行一个优化使O(N^2)转化为O(N*logN)的时间复杂度的算法（这个实现要求两个序列当中重复的元素比较少，这个题目两个序列没有重复的元素，所以没问题）。

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define MAXN 100100#define MAXE 5#define INF 1000000000#define MOD 10007#define LL long long#define ULL unsigned long long#define pi 3.14159using namespace std;int a[MAXN];int b[MAXN];int change[MAXN];int dp[MAXN];int arr[MAXN];int main() {    std::ios::sync_with_stdio(false);    int n, m, k, T;    cin >> T;    for (int kase = 1; kase <= T; ++kase) {        cin >> k >> n >> m;        n++, m++;        memset(change, -1, sizeof(change));        int cnt = 0;        for (int i = 1; i <= n; ++i) {            cin >> a[i];            change[a[i]] = i;        }        for (int i = 1; i <= m; ++i) {            cin >> b[i];            if (change[b[i]] != -1) {                dp[cnt++] = change[b[i]];            }        }        fill(arr, arr + MAXN, INF);        for (int i = 0; i < cnt; ++i) {            *lower_bound(arr, arr + cnt, dp[i]) = dp[i];        }        cout << "Case " << kase << ": " << lower_bound(arr, arr + cnt, INF) - arr << endl;    }    return 0;}/* 1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9 */