LCIS POJ 2172 Greatest Common Increasing Subsequence

 

题目传送门

题意：LCIS(Longest Common Increasing Subsequence) 最长公共上升子序列

分析：a[i] != b[j]: dp[i][j] = dp[i-1][j]; a[i]==b[j]:  dp[j]=max(dp[j],dp[k]); (1<=k<j&&b[k]<b[j])　　打印路径时按照b[i]来输出

收获：理解不是很深入，推荐资料: 最长公共上升子序列（LCIS）的O(n^2)算法　　最长公共上升子序列的另一个O(mn)的算法

 

 

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
using namespace std;

const int N = 5e2 + 10;
const int INF = 0x3f3f3f3f;
int a[N], b[N], dp[N][N], fx[N][N], fy[N][N];
int n, m;
bool fir;

void print(int x, int y, int last)	{		//bool fir;
	if (x == 0 || y == 0)	return ;
	print (fx[x][y], fy[x][y], y);
	if (y != last)	{
		if (fir)	printf ("%d", b[y]), fir = false;
		else	printf (" %d", b[y]);
	}
}

void LCIS(void)	{
	memset (dp, 0, sizeof (dp));
	memset (fx, 0, sizeof (fx));
	memset (fy, 0, sizeof (fy));
	int sx = 0, sy = 0;
	int ret = 0, k = 0;
	for (int i=1; i<=n; ++i)	{
		k = 0;
		for (int j=1; j<=m; ++j)	{
			dp[i][j] = dp[i-1][j];								//以a[]为主循环，每个a[i]，去找每个b[j]
			fx[i][j] = i - 1;	fy[i][j] = j;
			if (a[i] == b[j] && dp[i][j] < dp[i][k] + 1)	{		//满足LCS
				dp[i][j] = dp[i][k] + 1;						//在1~j-1找到b[k]<a[i]，满足LIS，在b[k]上更新dp
				fx[i][j] = i;	fy[i][j] = k;
			}
			else if (a[i] > b[j] && dp[i][j] > dp[i][k])	k = j;	//找到最优的k
			if (ret < dp[i][j])	{
				ret = dp[i][j];									//更新所有dp中的最大值
				sx = i, sy = j;
			}
		}
	}
	printf ("%d\n", ret);
	fir = true;
	print (sx, sy, -1);	puts ("");
}

int main(void)	{
	while (scanf ("%d", &n) == 1)	{
		for (int i=1; i<=n; ++i)	scanf ("%d", &a[i]);
		scanf ("%d", &m);
		for (int i=1; i<=m; ++i)	scanf ("%d", &b[i]);
		LCIS ();
	}

	return 0;
}