算法分析与实践-作业9

LCS算法

3. 设计

#include<stdio.h>
#include<string.h>
const int maxn = 1000 + 10;
char x[maxn], y[maxn];
int dp[maxn][maxn];
int b[maxn][maxn];   //b[i][j]=0为删除两个，b[i][j]=1为删除x，b[i][j]=2为删除y 
char lcs[maxn];
int n, m;             //n为字符串x的长度，m为字符串y的长度 
int main() {
    scanf("%d", &n);
    scanf("%s", x + 1);
    scanf("%d", &m);
    scanf("%s", y + 1);
    //初始化 
    for (int i = 0; i <= m; ++i)dp[0][i] = 0;
    for (int i = 0; i <= n; ++i)dp[i][0] = 0;

    /*----------以下为求最长公共子序列长度----------*/
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++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]) {
                    b[i][j] = 1;
                    dp[i][j] = dp[i - 1][j];
                }
                else {
                    b[i][j] = 2;
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
    }
    /*----------以上为求最长公共子序列长度----------*/
    /*----------以下为求最长公共子序列----------*/
    int i = n, j = m;
    while (i != 0 && j != 0) {
        if (b[i][j] == 0) {
            lcs[dp[i][j]] = x[i];
            i--, j--;
        }
        else if (b[i][j] == 1) {
            i--;
        }
        else {
            j--;
        }
    }
    /*----------以上为求最长公共子序列----------*/
    printf("最长公共子序列长度为：%d
", dp[n][m]);
    if (dp[n][m] != 0) {
        printf("最长公共子序列为：");
        for (int i = 1; i <= dp[n][m]; ++i) {
            printf("%c", lcs[i]);
        }
    }
}

4. 分析

T(n)=O(nm)

5. 源码

https://github.com/JayShao-Xie/algorithm-work/blob/master/LCS.cpp