HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1423

 

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

 题目大意是给定两个数字串seq1、seq2，求出它们最长公共递增子序列的长度。
状态 dp[j]表示seq1中从1到n与seq2中从1到j并以seq2[j]为结尾的最长公共上升子序列的长度。
状态转移方程： dp[j] = dp[k] + 1, if seq1[i] = seq2[j], 1 <= k < j.

#include <stdio.h>

#include <string.h>



#define MAX 501



int T;

int seq1[MAX], seq2[MAX];

int len1, len2;

int dp[MAX];



int LCIS(){

    int i, j;

    int Max;

    memset(dp, 0, sizeof(dp));

    for (i = 1; i <= len1; ++i){

        Max = 0;

        for (j = 1; j <= len2; ++j){

            if (seq1[i] > seq2[j] && Max < dp[j])

                Max = dp[j];

            if (seq1[i] == seq2[j])

                dp[j] = Max + 1;

        }

    }

    Max = 0;

    for (i = 1; i <= len2; ++i){

        if (Max < dp[i])

            Max = dp[i];

    }

    return Max;

}



int main(void){

    int i;

    scanf("%d", &T);

    while (T-- != 0){

        scanf("%d", &len1);

        for (i = 1; i <= len1; ++i)

            scanf("%d", &seq1[i]);

        scanf("%d", &len2);

        for (i = 1; i <= len2; ++i)

            scanf("%d", &seq2[i]);

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

        if (T)

            putchar('\n');

    }



    return 0;

}