实验9LCS算法

问题：

给定序列$X，Y$，求最长公共子序列。

解析：

设$dp[i][j]$表示前$i$个$x$和前$j$个$y$的最长公共子序列。

$dp[i][j]=max(dp[i][j],dp[i-1][j],dp[i][j-1])$  当前最长由前一个转移过来

$if(x[i]==y[j]) dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1)$   如果当前两位相等，最长序列长度$+1$

设计（核心代码）：

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 (x[i] == y[j]) {
            dp[i][j] = dp[i - 1][j - 1] + 1;
        }
        else {
            if (dp[i - 1][j] > dp[i][j - 1]) {
                del[i][j] = 1;
                dp[i][j] = dp[i - 1][j];
            }
            else {
                del[i][j] = 2;
                dp[i][j] = dp[i][j - 1];
            }
        }
    }
}

分析：

$O(n*m)$。

源码：

https://github.com/Big-Kelly/Algorithm

//#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 5e3 + 5;
const ll mod = 998244353;

string x, y, ans;
int len1, len2;
int dp[maxn][maxn], del[maxn][maxn];

int main() {
    cin >> x >> y;
    len1 = x.length(), len2 = y.length();
    x = " " + x, y = " " + y;
    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 (x[i] == y[j]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            }
            else {
                if (dp[i - 1][j] > dp[i][j - 1]) {
                    del[i][j] = 1;
                    dp[i][j] = dp[i - 1][j];
                }
                else {
                    del[i][j] = 2;
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
    }
    int i = len1, j = len2;
    ans = "";
    while (i && j) {
        if (!del[i][j]) {
            ans = x[i] + ans;
            i--, j--;
        }
        else if (del[i][j] == 1) {
            i--;
        }
        else {
            j--;
        }
    }
    cout << "LCS长度:" << dp[len1][len2] << endl;
    cout << "LCS:" << ans << endl;
}