hdoj1423 Greatest Common Increasing Subsequence

hdoj1423 Greatest Common Increasing Subsequence

时空限制    1000ms/64MB

【问题描述】

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequenc

【输入格式】

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

【输出格式】

output print L - the length of the greatest common increasing subsequence of both sequences.

【样例输入】

1

5
1 4 2 5 -12
4
-12 1 2 4

【样例输出】

2

 

【代码】

f[i][j]表示以数组a的前i个元素与数组b的前j个元素且以b[j]为结尾构成的LCIS的长度

f[i][j] = f[i-1][j] (a[i] != b[j])

f[i][j] = max(f[i-1][k]+1) (1 <= k < j && b[j] > b[k]) (a[i]==b[j])

法一：O(n^3)  空间n^2

#include	//O(n^3)
#include
using namespace std;
const int N = 505;
int T,n,m,a[N],b[N],f[N][N];

int main(){
	cin>>T;
	while (T--){
		cin>>n;
		for (int i=1; i<=n; i++) cin>>a[i];
		cin>>m;
		for (int i=1; i<=m; i++) cin>>b[i];
		fill(f[0],f[0]+N*N,0);
		for (int i=1; i<=n; i++)
			for (int j=1; j<=m; j++)
				if (b[j]!=a[i]) f[i][j]=f[i-1][j];
				else {
					for (int k=1; k

法二：O(n^2)  空间n^2

维护一个maxv值来储存最大的f[i-1][k]值。即只要有b[j]

#include	//O(n^2)
#include
using namespace std;
const int N = 505;
int T,n,m,a[N],b[N],f[N][N];

int main(){
	cin>>T;
	while (T--){
		cin>>n;
		for (int i=1; i<=n; i++) cin>>a[i];
		cin>>m;
		for (int i=1; i<=m; i++) cin>>b[i];
		fill(f[0],f[0]+N*N,0);
		for (int i=1; i<=n; i++){
			int maxv=0;
			for (int j=1; j<=m; j++){
				if (b[j]

法三：O(n^3)  空间n

上面的代码有些地方与0/1背包很相似，即每次用到的只是上一层循环用到的值，即f[i-1][j]，那么我们可以像优化0/1背包问题利用滚动数组来优化空间。

#include	//O(n^2) 空间优化O(n)
#include
using namespace std;
const int N = 505;
int T,n,m,a[N],b[N],f[N];

int main(){
	cin>>T;
	while (T--){
		cin>>n;
		for (int i=1; i<=n; i++) cin>>a[i];
		cin>>m;
		for (int i=1; i<=m; i++) cin>>b[i];
		fill(f,f+N,0);
		for (int i=1; i<=n; i++){
			int maxv=0;
			for (int j=1; j<=m; j++){
				if (b[j]