POJ 题目2127 Greatest Common Increasing Subsequence（LICS，输出路径）

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本文介绍了一个经典的计算机科学问题——寻找两个整数序列之间的最长公共递增子序列，并提供了一段C语言实现的代码示例。该问题旨在通过算法找出两个序列中相同且递增的最长子序列。

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Greatest Common Increasing Subsequence

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 9762		Accepted: 2584
Case Time Limit: 2000MS		Special Judge

Description

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
Sequence S ₁ , S ₂ , . . . , S _N of length N is called an increasing subsequence of a sequence A ₁ , A ₂ , . . . , A _M of length M if there exist 1 <= i ₁ < i ₂ < . . . < i _N <= M such that S _j = A _{i_j} for all 1 <= j <= N , and S _j < S _j+1 for all 1 <= j < N .

Input

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

Output

On the first line of the output file print L --- the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

Sample Input

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

Sample Output

2
1 4

Source

Northeastern Europe 2003, Northern Subregion

ac代码

#include<stdio.h>
#include<string.h>
int n,m;
int dp[1010][1010],path[1010][1010],ans[1010],aj,ai,res,a[1010],b[1010];
void LCS()
{
	int i,j,mj;
	res=0;
	memset(dp,0,sizeof(dp));
	memset(path,-1,sizeof(path));
	for(i=1;i<=n;i++)
	{
		int ma=0;
		for(j=1;j<=m;j++)
		{
			dp[i][j]=dp[i-1][j];
			if(b[j]<a[i]&&dp[i][j]>ma)
			{
				ma=dp[i][j];
				mj=j;
			}
			else
			{
				if(b[j]==a[i])
				{
					dp[i][j]=ma+1;
					path[i][j]=mj;
				}
			}
			if(res<dp[i][j])
			{
				res=dp[i][j];
				ai=i;
				aj=j;
			}
		}
	}
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		int i,j;
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		scanf("%d",&m);
		for(j=1;j<=m;j++)
			scanf("%d",&b[j]);
		LCS();
		printf("%d\n",res);
		int temp=res;
		while(temp)
		{
			if(path[ai][aj]>-1)
			{
				ans[temp--]=b[aj];
				aj=path[ai][aj];
			}
			ai--;
		}
		for(i=1;i<=res;i++)
			printf("%d%c",ans[i],i==res?'\n':' ');
	}
}