题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1423
1 5 1 4 2 5 -12 4 -12 1 2 4
2
#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;
}HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划,布布扣,bubuko.com
HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划
原文:http://blog.csdn.net/jdplus/article/details/20131931