算法分析设计实践——最长公共子序列

算法分析设计实践——最长公共子序列

1.问题

对于序列a和序列b，求其最长公共子序列

2.解析

通过动态规划的方式

dp[i][j] 前i个字符的x和前j个字符的y的最长公共子序列

当a[i] = b[j] 的时候

dp[i][j] = max(dp[i][j] , dp[i - 1][j - 1] + 1)

如果第i位和第j位相等的话，那么最长序列长度 + 1

当 a[i] != b[j] 的时候

dp[i][j] = max(dp[i][j] , dp[i - 1][j] , dp[i][j - 1])

如果第i位和第j不相等的话 ， 那么最长公共子序列的长度则从 dp[i - 1][j] 和 dp[i][j - 1] 转移过来

3.设计

void lcs()
{
    for (int i = 0; i <= len1; ++i) dp[i][0] = 0;
    for (int i = 0; i <= len2; ++i) dp[0][i] = 0;
    for (int i = 1; i <= len1; ++i)
    {
        for (int j = 1; j <= len2; ++j)
        {
            if (arr[i] == brr[j])
                dp[i][j] = dp[i - 1][j - 1] + 1;
            else
            {
                if (dp[i - 1][j] > dp[i][j - 1])
                {
                    res[i][j] = 1;
                    dp[i][j] = dp[i - 1][j];
                }
                else
                {
                    res[i][j] = 2;
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
    }
}
string getlcs()
{
    int i = len1, j = len2;
    string ans = "";
    while (i > 0 && j > 0)
    {
        if (!res[i][j])
        {
            ans = arr[i] + ans;
            i--, j--;
        }
        else if (res[i][j] == 1)
        {
            i--; 
        }
        else
        {
            j--;
        }
    }
    return ans;
}

4.分析

时间复杂度：O(n * m)

5.源码

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<vector>
#include<queue>
#include<set>
#include<stack>
#include<map>
#include<cctype>
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid + 1,r
#define P pair<int,int>
#define ull unsigned long long
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
const ll mod = 998244353;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;

inline ll read()
{
    ll X = 0, w = 0; char ch = 0;
    while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}

int len1, len2;
int dp[1000][1000];
string arr, brr;
int res[1000][1000];
void lcs()
{
    for (int i = 0; i <= len1; ++i) dp[i][0] = 0;
    for (int i = 0; i <= len2; ++i) dp[0][i] = 0;
    for (int i = 1; i <= len1; ++i)
    {
        for (int j = 1; j <= len2; ++j)
        {
            if (arr[i] == brr[j])
                dp[i][j] = dp[i - 1][j - 1] + 1;
            else
            {
                if (dp[i - 1][j] > dp[i][j - 1])
                {
                    res[i][j] = 1;
                    dp[i][j] = dp[i - 1][j];
                }
                else
                {
                    res[i][j] = 2;
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
    }
}
string getlcs()
{
    int i = len1, j = len2;
    string ans = "";
    while (i > 0 && j > 0)
    {
        if (!res[i][j])
        {
            ans = arr[i] + ans;
            i--, j--;
        }
        else if (res[i][j] == 1)
        {
            i--; 
        }
        else
        {
            j--;
        }
    }
    return ans;
}

int main()
{
    cin >> arr >> brr;
    len1 = arr.size(), len2 = brr.size();
    arr = " " + arr;
    brr = " " + brr;
    lcs();
    cout << "最长公共字串长度： " << dp[len1][len2] << endl;
    cout << "最长公共字串： " << getlcs() << endl;
    return 0;
}

https://github.com/BambooCertain/Algorithm.git