5 字符串拼接

5 字符串拼接作者: 赵晓鹏时间限制: 1S章节: 动态规划与贪心

输入说明 :

见问题描述。

输出说明 :

见问题描述。

输入范例 :

aaaaaaaaaab
aaaaaaaac
aaaaaaaaaacaaaaaaaab

输出范例 :

YES
aa_________aaaaaaaab

 

#include
#include
using namespace std;
bool dp[6000][6000];
    
bool isInterleave(string s1, string s2, string s3) {
    int len1 = s1.length();
    int len2 = s2.length();
    int len3 = s3.length();

    if (len1 + len2 != len3)return false;
    //表示s1[0..i]和s2[0..j]能否组成s3[0..i+j]



    dp[0][0] = true;
    //初始化
    for (int i = 1; i <= len1; i++)
    {
        dp[i][0] = dp[i - 1][0] &&s1[i - 1] == s3[i - 1];
    }

    for (int j = 1; j <= len2; j++)
    {
        dp[0][j] = dp[0][j - 1] && s2[j - 1] == s3[j - 1];
    }
            for (int i = 1; i <= len1; i++) {
                for (int j = 1; j <= len2; j++) {
                    //内层成立再判断当前是否相等
                    dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1] ||
                        dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];
                }
            }
            return dp[len1][len2];
        }
       

int main() {
    string s1;
    string s2;
    string s3;

    cin >> s1;
    cin >> s2;
    cin >> s3;

    bool judge=isInterleave(s1, s2, s3);
    if (judge = true) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO";
    }



    string result = "";
    int y_count = 0;
    int x_count = 0;
    int x_length = s1.length();
    int y_length = s2.length();

    int x_now = 0;
    int y_now = 0;
    int x_next = 0;
    int y_next = 0;
    int x_pre = 0;
    int y_pre = 0;

    x_now = x_length;
    y_now = y_length;

    bool statue = 0;//0默认为左边走
    if (dp[x_now][y_now] == 1)
    {
        while (1) 
        {
            if (x_now == 0 && y_now == 0) {
                break;
            }


            if (statue == 0)//默认往左边走
            {
                if (dp[x_now ][y_now-1] == 1)
                {
                    result = "_" + result;
                    y_now = y_now - 1;//更新位置
                    y_count++;
              
                }

                else if (dp[x_now ][y_now-1] == 0)
                {
                    statue = 1;//切换方向
                    result = s1[x_length - x_count - 1] + result;
                    x_now--;
                    x_count++;
               
                }
            }

            else if (statue == 1) 
            {
                if (dp[x_now - 1][y_now] == 1 && dp[x_now][y_now - 1] == 1) 
                {
                    //也要切换状态
                    statue == 0;//切换状态
                    result = "_" + result;
                    y_now--;
                    y_count++;
                   
                }
                else if (dp[x_now-1][y_now ] == 1)
                {
                    result = s1[x_length - x_count - 1] + result;
                    x_now--;
                    x_count++;
              
                }
                else if (dp[x_now-1][y_now] == 0) {
                    statue == 0;//切换状态
                    result = "_" + result;
                    y_now--;
                    y_count++;
                    
                }

            }
        }
    }


    cout << result << endl;
    return 0;
}