HDOJ-----1159Common Subsequence

最新推荐文章于 2020-04-13 10:58:43 发布

pupil_blue

最新推荐文章于 2020-04-13 10:58:43 发布

阅读量239

点赞数

CC 4.0 BY-SA版权

分类专栏： LIS/LCS

本文链接：https://blog.youkuaiyun.com/blue_hpu/article/details/52197553

LIS/LCS 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种解决最长公共子序列问题的经典算法实现。通过使用滚动数组的方式优化内存消耗，该算法能有效地找出两个字符串之间的最长公共子序列长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34248 Accepted Submission(s): 15619

Problem Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

最长公共子序列，模板题

滚动数组可以减少点内存消耗

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 10010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
char st[maxn], str[maxn];
int dp[2][maxn];
int main(){
	int ans, n, l1, l2;
	while(scanf("%s%s", st, str) != EOF){
        l1 = strlen(st);
        l2 = strlen(str);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= l1; i++){
            for(int j = 1; j <= l2; j++){
                if(st[i-1] == str[j-1]){
                    dp[i&1][j] = dp[(i-1)&1][j-1]+1;
                }
                else{
                    dp[i&1][j] = max(dp[(i-1)&1][j], dp[i&1][j-1]);
                }
            }
        }
        printf("%d\n", dp[l1&1][l2]);
	}
	return 0;
}