10405 - Longest Common Subsequence UVA

最新推荐文章于 2020-01-27 10:07:07 发布

Just_Try

最新推荐文章于 2020-01-27 10:07:07 发布

阅读量443

点赞数

CC 4.0 BY-SA版权

分类专栏： UVA 动态规划

本文链接：https://blog.youkuaiyun.com/abc123456789cbaabc/article/details/8661694

动态规划同时被 2 个专栏收录

25 篇文章

订阅专栏

UVA

20 篇文章

订阅专栏

本文介绍了一种求解两个字符串最长公共子序列的问题，并通过动态规划的方法给出了详细的算法实现过程。文章提供了完整的C++代码示例，展示了如何利用递归与数组两种方式高效地解决此类问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Longest Common Subsequence

Sequence 1:                
Sequence 2:                Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:
abcdgh
aedfhr

is adh of length 3.

Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

自己做的第一个动态规划的基础题：

关键是找到其状态转移方程！！！

不妨设d（x，y）代表字符串A1，A2....Ax和字符串B1,B2.....By

最长公共字符串

那么当Ax=By的时候

有：

d（x，y）= d(x-1,y-1)+1;
如果Ax ！=By时
那么：
d(x,y) = max(d(x-1,y),d(x,y-1);
代码如下：
A[x-1]==B[y-1]//注意是x-1和y-1而非x，y

注意读入字符串此处必须用gets用scanf会wr

杭电完全类似的题目

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
char A[1100];
char B[1100];
int d[1100][1100];
int vis[1100][1100];
int dp(int x,int y)
{
    if(vis[x][y])
    return d[x][y];
    vis[x][y] = 1;
    if(x==0||y==0)
    d[x][y] = 0;
    else
    {
    if(A[x-1]==B[y-1])
     d[x][y] = dp(x-1,y-1) + 1;
    else
     d[x][y] = max(dp(x-1,y),dp(x,y-1));
    }
     return d[x][y];
}
int main()
{
    while(gets(A)&&gets(B))
    {
        memset(vis,0,sizeof(vis));
        int la = strlen(A);
        int lb = strlen(B);
        for(int i = 0;i<=lb;++i)
          d[0][i] = 0;
        for(int i = 0;i<=la;++i)
          d[i][0] = 0;
          printf("%d\n",dp(la,lb));
    }
}

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
char A[1100];
char B[1100];
int d[1100][1100];
int vis[1100][1100];
int dp(int x,int y)
{
    if(vis[x][y])
    return d[x][y];
    vis[x][y] = 1;
    if(x==0||y==0)
    d[x][y] = 0;
    else
    {
    if(A[x-1]==B[y-1])
     d[x][y] = dp(x-1,y-1) + 1;
    else
     d[x][y] = max(dp(x-1,y),dp(x,y-1));
    }
     return d[x][y];
}
int main()
{
    while(gets(A)&&gets(B))
    {
        memset(vis,0,sizeof(vis));
        int la = strlen(A);
        int lb = strlen(B);
          printf("%d\n",dp(la,lb));
    }
}

下面是完全用数组做的：

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
char A[1100];
char B[1100];
int d[1100][1100];
int main()
{
    while(gets(A)&&gets(B))
    {
        int la = strlen(A);
        int lb = strlen(B);
        for(int i = 0;i<=la;++i)
        d[i][0] = 0;
        for(int i = 0;i<=lb;++i)
        d[0][i] = 0;
        for(int i = 1;i<=la;++i)
          for(int j = 1;j<=lb;++j)
          {
              if(A[i-1]==B[j-1])
              d[i][j] = d[i-1][j-1]+1;
              else
              d[i][j] = max(d[i-1][j],d[i][j-1]);
          }
        printf("%d\n",d[la][lb]);
    }
}