Subsequence
题目背景：
1.10 WC模拟T3
分析：DP + 复杂度分析
这个题感觉难点在复杂度分析，看似总状态数n * m转移复杂度k，但是实际上，可以证明得到答案数与k的乘积是小于(n + m)的，证明：假设当前匹配到A的第j位，B的第i位，那么第j + 1位到第j + k - 1位，那么一定有一个数没有出现过，那么选择这个数就会让i + j增加k，所以每一次操作至少能增加k，所以最后的答案不超过(n + m) / k，所以直接按照O(n3)的方式DP就可以了，实际复杂度为O(n2)，定义dp[i][j]表示，当前枚举第i位，在A串中匹配到j，dp[i][j]表示在B串中匹配到的最后的位置。预处理next_a[i][j]表示第i位之后第一个j出现的位置，最后结尾设为n + 1，那么当dp[i][n + 1] = m + 1时，说明i就是可行的长度了。
Source:
/*created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;
}///*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = read(), iosig = false; !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true; }for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;
}template<class T>
inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}
}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);
}/*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = getchar(), iosig = false; !isdigit(c); c = getchar())if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int MAXN = 4000 + 10;int n, m, k;
int dp[MAXN][MAXN], a[MAXN], b[MAXN];
int next_a[MAXN][MAXN], next_b[MAXN][MAXN];inline void read_in() {R(n), R(m), R(k);for (int i = 1; i <= n; ++i) R(a[i]);for (int i = 1; i <= m; ++i) R(b[i]);for (int i = 1; i <= k; ++i) {next_a[n][i] = next_a[n + 1][i] = n + 1;next_b[m][i] = next_b[m + 1][i] = m + 1;}for (int i = n - 1; i >= 0; --i) {for (int j = 1; j <= k; ++j)next_a[i][j] = next_a[i + 1][j];next_a[i][a[i + 1]] = i + 1;}for (int i = m - 1; i >= 0; --i) {for (int j = 1; j <= k; ++j)next_b[i][j] = next_b[i + 1][j];next_b[i][b[i + 1]] = i + 1;}
}inline void solve() {memset(dp, -1, sizeof(dp));dp[0][0] = 0;for (int i = 0; ; ++i) {for (int j = 0; j <= n + 1; ++j) {if (dp[i][j] == -1) continue ;if (j == n + 1 && dp[i][j] == m + 1) std::cout << i, exit(0);for (int c = 1; c <= k; ++c)dp[i + 1][next_a[j][c]] = std::max(dp[i + 1][next_a[j][c]], next_b[dp[i][j]][c]);}}
}int main() {freopen("subsequence.in", "r", stdin);freopen("subsequence.out", "w", stdout);read_in();solve();return 0;
}