hdu5904 LCIS dp

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LCIS

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5 Accepted Submission(s): 300

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

参考：

看了看别人的题解，自己码一遍，感觉好简单……为什么自己就是做不出来……

a[x]为第一个序列中到x为止最长连续上升子序列长度，b[x]为第二个序列中到x为止最长的连续上升子序列长度，均包括x

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 100000;

int T, n, m, x;
int a[maxn + 5], b[maxn + 5];

int main()
{
	cin >> T;
	while (T--) {
		scanf("%d%d", &n, &m);
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		for (int i = 0; i < n; i++) {
			scanf("%d", &x);
			a[x] = max(a[x], a[x - 1] + 1); //因为是一个方向扫，所以max可以不用写也对
		}
		int ans = -1;
		for (int i = 0; i < m; i++) {
			scanf("%d", &x);
			b[x] = max(b[x], b[x - 1] + 1);
			//min相当于取交集，例如1, 2, 3   a[3] = 3;  2, 3  b[3] = 2，那么肯定取min， 即2, 3
			ans = max(ans, min(a[x], b[x]));
		}
		printf("%d\n", ans);
	}
	return 0;
}

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