Greatest Common Increasing Subsequence HDU - 1423

问题：

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

大意：

给你两个序列，求这两个序列的最长公共上升子序列的长度。

思路：

这就相当于是求最长公共序列和最长上升序列的合并。

状态迁移方程：dp[ i+1 ][ j+1 ]=max( dp[ i ][ j ]+1, dp[ i+1 ][ j ] ,dp[ i ][ j+1 ]),

代码：

#define N 510
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
int k[N],s[N];
int kl,sl;
int dp[N][N];//设dp[i+1][j+1]是第一个序列前i个与第二个序列前j个的最长公共上升子序列
long long int d[N][N];//保留每段以求出序列末尾的最小值
int main()
{
    int c;scanf("%d",&c);
    while(c--)
    {
        scanf("%d",&kl);
        for(int i=0; i<kl; i++)
            scanf("%d",&k[i]);
        scanf("%d",&sl);
        for(int i=0; i<sl ;i++)
            scanf("%d",&s[i]);
        memset(d,-0x3f,sizeof(d));
        for(int i=0; i<kl; i++)
            for(int j=0; j<sl; j++)
            {
                if(k[i]==s[j]&&d[i][j]<k[i])//序列可以延长
                {
                    dp[i+1][j+1]=dp[i][j]+1;
                    d[i+1][j+1]=k[i];//保留最小值
                }
                else  //不能延长
                {
                    if(dp[i+1][j]>dp[i][j+1])//谁长留下谁
                    {
                        dp[i+1][j+1]=dp[i+1][j];
                        d[i+1][j+1]=d[i+1][j];
                    }
                    else if(dp[i+1][j]<dp[i][j+1])
                    {
                        dp[i+1][j+1]=dp[i][j+1];
                        d[i+1][j+1]=d[i][j+1];
                    } 
                    else //一样长留下，序列末尾值最小的一个
                    {
                        dp[i+1][j+1]=dp[i+1][j];
                        d[i+1][j+1]=min(d[i+1][j],d[i][j+1]);
                    }
                }
            }
        printf("%d\n",dp[kl][sl]);
        if(c)printf("\n");
    }
    return 0;
}