Problem Description
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.
Input
Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.
Output
output print L - the length of the greatest common increasing subsequence of both sequences.
Sample Input
1 5 1 4 2 5 -12 4 -12 1 2 4
Sample Output
2
Source
ACM暑期集训队练习赛(二)
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求最长公共上升子序列
设dp[i][j] 表示a的前i个数和b的前j个数,以b[j]结尾的LCIS长度
如果a[i] != b[j]
令dp[i][j] = dp[i - 1][j]; //即取上一次的最优解
如果a[i] == b[j]
那么dp[i][j] = max (dp[i - 1][k]) + 1;
加以优化,复杂度可以控制在O(m*n)
/*************************************************************************
> File Name: hdu1423.cpp
> Author: ALex
> Created Time: 2015年02月21日 星期六 19时12分48秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int N = 510;
int dp[N][N];
int a[N];
int b[N];
int main ()
{
int t;
scanf("%d", &t);
while (t--)
{
int n, m;
scanf("%d", &n);
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]);
}
memset (dp, 0, sizeof(dp));
for (int i = 1; i <= n; ++i)
{
int maxs = 0;
for (int j = 1; j <= m; ++j)
{
dp[i][j] = dp[i - 1][j];
if (a[i] > b[j] && maxs < dp[i][j])
{
maxs = dp[i][j];
}
if (a[i] == b[j])
{
dp[i][j] = maxs + 1;
}
}
}
int ans = 0;
for (int i = 1; i <= m; ++i)
{
ans = max (dp[n][i], ans);
}
printf("%d\n", ans);
if (t)
{
printf("\n");
}
}
return 0;
}