WC模拟（1.10） T3 Subsequence

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;
}