BestCoder Round #87（1003-【思维】【LIS&&LCS】）

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探讨了如何求解两个序列中最长的共增子序列问题，通过动态规划方法找到以每个元素结尾的最大子序列长度。

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题目链接：点击打开链接

1003-LCIS

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 243 Accepted Submission(s): 98

Problem Description

Alex has two sequences

a1,a2,...,an and

b1,b2,...,bm . He wants find a longest common subsequence that consists of consecutive values in increasing order.

Input

There are multiple test cases. The first line of input contains an integer

T , indicating the number of test cases. For each test case:

The first line contains two integers

n and

(1≤n,m≤100000) -- the length of two sequences. The second line contains

n integers:

a1,a2,...,an

(1≤ai≤106) . The third line contains

n integers:

b1,b2,...,bm

(1≤bi≤106) .

There are at most

1000 test cases and the sum of

n and

m does not exceed

2×106 .

Output

For each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.

Sample Input

  
   3
3 3
1 2 3
3 2 1
10 5
1 23 2 32 4 3 4 5 6 1
1 2 3 4 5
1 1
2
1

Sample Output

Source

BestCoder Round #87

题解：

令 $f (i)$ 是以 $a_i$ 结尾的最大值, $g (i)$ 是以 $b_i$ 结尾的最大值. 答案就是 $\max_{a_i = b_j} {\min(f(i),g(j)}$ . $f$ 和 $g$ 随便dp一下就出来了.

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN=1e5+10;
int n,m;
int a[MAXN],b[MAXN];
int dp1[MAXN],dp2[MAXN];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&m);
		for(int i=0;i<MAXN;i++) 
		{
			a[i]=b[i]=dp1[i]=dp2[i]=0;
		}
//		这里用 for来初始化数组也可以，用 memset一般初始化一些较大的结构体和数组比较省时
//		像空间比较小的如 100，用 for就比较快
//		memset(a,0,sizeof(a));
//		memset(b,0,sizeof(b));
//		memset(dp1,0,sizeof(dp1));
//		memset(dp2,0,sizeof(dp2));
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			dp1[a[i]]=max(dp1[a[i]],dp1[a[i]-1]+1);
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d",&b[i]);
			dp2[b[i]]=max(dp2[b[i]],dp2[b[i]-1]+1);
		}
		int ans=0;
		for(int i=0;i<n;i++)
		{
			ans=max(ans,min(dp1[a[i]],dp2[a[i]]));
		}
//		for(int i=0;i<m;i++) 任意找一个(a或 b) dp都可以 
//		{
//			ans=max(ans,min(dp1[b[i]],dp2[b[i]]));
//		} 
		printf("%d\n",ans);
	}
	return 0;
}