【14.06%】【hdu 5904】LCIS

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

Total Submission(s): 519 Accepted Submission(s): 238

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 m (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】

设q[i],表示以数字i为上升序列的最后一个元素的最长长度。

则if (q[i-1] == 0)

q[i-1] = 1;

else

q[i] = max(q[i-1]+1,q[i]);

最后枚举要枚举n+m个而不是枚举1..100W,不然会超时。

【代码】

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

const int MAXN = 102001;

const int MAX_NUM = 1009999;

int n,m,a[MAXN],b[MAXN],q1[MAX_NUM],q2[MAX_NUM];

void input_data()

{

	scanf("%d%d", &n,&m);

	for (int i = 1; i <= n; i++)

	{

		scanf("%d", &a[i]);

		if (q1[a[i]-1] == 0)

			q1[a[i]] = 1;

		else

			if (q1[a[i]] < q1[a[i]-1]+1)

				q1[a[i]] = q1[a[i] - 1] + 1;

	}

	for (int i = 1; i <= m; i++)

	{

		scanf("%d", &b[i]);

		if (q2[b[i] - 1] == 0)

			q2[b[i]] = 1;

		else

			if (q2[b[i]] < q2[b[i] - 1] + 1)

				q2[b[i]] = q2[b[i] - 1] + 1;

	}

}

void get_ans()

{

	int len = 0;

	for (int i = 1; i <= n; i++)//枚举的是出现过的每个数字

	{

		int d = min(q1[a[i]], q2[a[i]]);

		if (d > len)

			len = d;

	}

	for (int i = 1; i <= m; i++)

	{

		int d = min(q1[b[i]], q2[b[i]]);

		if (d > len)

			len = d;

	}

	printf("%d\n", len);

}

void init()

{

	for (int i = 0; i <= 1000000; i++)

		q1[i] = 0;

	for (int i = 0; i <= 1000000; i++)

		q2[i] = 0;

}

int main()

{

//	freopen("F:\\rush.txt", "r", stdin);

	int t;

	scanf("%d", &t);

	while (t--)

	{

		init();

		input_data();

		get_ans();

	}

	return 0;

}

巴特西

【14.06%】【hdu 5904】LCIS

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