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
", LCIS());
        if (T)
            putchar('
');
    }

    return 0;
}

HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划

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