題目要你按字典序輸出兩個字串的多個最長公共子序列

思路：

先用動態規劃求兩個字串的最長公共子序列的儲存在dp[i][j];dp[i][j]表示s1字串1到i和s2字串1到j的最長公共子序列的長度

然後用兩個變數last1[i][j],last2[i][j]來分別儲存字元j(a的序號為0，b的序號為1，.....z的序號為25)在字串1-i中出現的最大標號，要是字元j沒有出現,則last[i][j]= 0;

然後從兩個字串的長度len1和len2開始列舉a---z字元,比如開始 t1 = last1[len1][0], t2 = last2[len2][0]表示a在s1字串1---len1的最大下標為

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <set>
using namespace std;

int const maxn = 100;
char s1[maxn], s2[maxn], tmp[maxn];
int dp[maxn][maxn], last1[100][30], last2[100][30], len1, len2, cnt;
set<string> collection;

void LCA();
void handle(char s[], int len, int last[][30]);
void find(int index1, int index2, int len);

int main()
{
   
    scanf("%s %s", &s1[1], &s2[1]);
    LCA();
    
    handle(s1, len1, last1);
    handle(s2, len2, last2);
    
    cnt = dp[len1][len2];
    tmp[cnt+1] = '\0';
    find(len1, len2, cnt);
    for(set<string>::iterator iter = collection.begin(); iter != collection.end(); iter++)
        printf("%s\n", iter->c_str());
    
    return 0;
}

void LCA()
{
    int i, j;
    for(i = 1; s1[i] != '\0'; i++)
    {
        for(j = 1; s2[j] != '\0'; j++)
        {
            if(s1[i] == s2[j])
                dp[i][j] = dp[i-1][j-1] + 1;
            else if(dp[i-1][j] >= dp[i][j-1])
                dp[i][j] = dp[i-1][j];
            else
                dp[i][j] = dp[i][j-1];
        }
    }
    len1 = i - 1;
    len2 = j - 1;
}

void handle(char s[], int len, int last[][30])
{
    for(int i = 0; i < 26; i++)
    {
        char c = 'a' + i;
        for(int j = 1; j <= len; j++)
        {
            int k;
            for(k = j; k >= 1; k--)
            {
                if(c == s[k])
                {
                    last[j][i] = k;
                    break;
                }
            }
            
            if(k == 0)
                last[j][i] = 0;
        }
    }
}

void find(int index1, int index2, int len)
{
    if(len <= 0)
    {
        collection.insert(&tmp[1]);
        return;
    }
    
    if(index1 > 0 && index2 > 0)
    {
        for(int i = 0; i < 26; i++)
        {
            int t1 = last1[index1][i];
            int t2 = last2[index2][i];
            if(dp[t1][t2] == len)
            {
                tmp[len] = 'a' + i;
                find(t1 - 1, t2 - 1, len - 1);
            }
        }
    }
}