LCIS 滚动数组 + O (N^2)

前面介绍了LCIS的两种不同时间复杂度的解法，但是空间复杂度同样都是O(N * N)，对于一些题目会存在空间超限的可能性，因此可以利用滚动数组对空间进行优化。

如果不知道LCIS的两种解法 --> O(N * N) --> O(N ^ 3)

观察下面的状态转移方程的代码：这其实是O(N ^ 2)的解法的代码

for (int i = 1; i <= n; i++){
	int val = 0;
	if(b[0] < a[i])val = dp[i - 1][0];
	for (int j = 1; j <= n; j++){
		if(a[i] == b[j])dp[i][j] = val + 1;
		else dp[i][j] = dp[i - 1][j];
		if(b[j] < a[i])val = max(val, dp[i - 1][j]);
	}
}

我们可以看出，状态转移的过程中，只用到了上一层（dp[ i - 1] [ j ]的状态），因此可以利用滚动数组对空间进行优化，用同一个数组接连保存后继的每一个状态，后面的覆盖前面的即可

int val = 0;
if(b[0] < a[i])val = dp[1][0];
for (int j = 1; j <= n; j++){
	if(a[i] == b[j])dp[2][j] = val + 1;
	else dp[i][j] = dp[1][j];
	if(b[j] < a[i])val = max(val, dp[1][j]);
}
for (int j = 1; j <= n; j++){
	dp[1][j] = dp[2][j];
}

完整代码

#include <bits/stdc++.h>
using namespace std;
const int N = 3010;
int a[N], b[N], dp[N][N], n;
template <class T>
bool read(T & a){
	a = 0;
	int flag = 0;
	char ch;
	if((ch = getchar()) == '-'){
		flag = 1;
	}
	else if(ch >= '0' && ch <= '9'){
		a = a * 10 + ch - '0';
	}
	while ((ch = getchar()) >= '0' && ch <= '9'){
		a = a * 10 + ch - '0';
	}
	if(flag)a = -a;
	return true;
}
template <class T, class ... R>
bool read(T & a, R & ... b){
	if(!read(a))return 0;
	read(b...);
}
template <class T>
bool out(T a){
	if(a < 0)putchar('-');
	if(a >= 10)out(a / 10);
	putchar(a % 10 + '0');
	return true;
}
template <class T, class ... R>
bool out(T a, R ... b){
	if(!out(a))return 0;
	out(b...);
}
int main()
{
	read(n);
	for (int i = 1; i <= n; i++)read(a[i]);
	for (int i = 1; i <= n; i++)read(b[i]);
	for (int i = 1; i <= n; i++){
		int val = 0;
//		if(b[0] < a[i])val = dp[i - 1][0];
		if(b[0] < a[i])val = dp[1][0];
//		for (int j = 1; j <= n; j++){
//			if(a[i] == b[j])dp[i][j] = val + 1;
//			else dp[i][j] = dp[i - 1][j];
//			if(b[j] < a[i])val = max(val, dp[i - 1][j]);
//		}
		for (int j = 1; j <= n; j++){
			if(a[i] == b[j])dp[2][j] = val + 1;
			else dp[i][j] = dp[1][j];
			if(b[j] < a[i])val = max(val, dp[1][j]);
		}
		for (int j = 1; j <= n; j++){
			dp[1][j] = dp[2][j];
		}
	}
	int res = 0;
//	for (int i = 1; i <= n; i++)res = max(res, dp[n][i]);
	for (int i = 1; i <= n; i++)res = max(res, dp[2][i]);
	printf("%d\n", res);
	return 0;
}