HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划

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

[align=left]Problem Description[/align]
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

[align=left]Input[/align]
Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

[align=left]Output[/align]
output print L - the length of the greatest common increasing subsequence of both sequences.

[align=left]Sample Input[/align]

1

5
1 4 2 5 -12
4
-12 1 2 4

[align=left]Sample Output[/align]

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;
}