Compromise————LCS+输出路径

﹏ヽ暗。殇╰゛Y 2022-05-17 10:08 144阅读 0赞

In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germany will fulfill the criteria, our government has so many wonderful options (raise taxes, sell stocks, revalue the gold reserves,…) that it is really hard to choose what to do.

Therefore the German government requires a program for the following task:  
Two politicians each enter their proposal of what to do. The computer then outputs the longest common subsequence of words that occurs in both proposals. As you can see, this is a totally fair compromise (after all, a common sequence of words is something what both people have in mind).

Your country needs this program, so your job is to write it for us.  
**Input**  
The input will contain several test cases.  
Each test case consists of two texts. Each text is given as a sequence of lower-case words, separated by whitespace, but with no punctuation. Words will be less than 30 characters long. Both texts will contain less than 100 words and will be terminated by a line containing a single ‘\#’.  
Input is terminated by end of file.  
**Output**  
For each test case, print the longest common subsequence of words occuring in the two texts. If there is more than one such sequence, any one is acceptable. Separate the words by one blank. After the last word, output a newline character.  
**Sample Input**  
die einkommen der landwirte  
sind fuer die abgeordneten ein buch mit sieben siegeln  
um dem abzuhelfen  
muessen dringend alle subventionsgesetze verbessert werden  
\#  
die steuern auf vermoegen und einkommen  
sollten nach meinung der abgeordneten  
nachdruecklich erhoben werden  
dazu muessen die kontrollbefugnisse der finanzbehoerden  
dringend verbessert werden  
\#  
**Sample Output**  
die einkommen der abgeordneten muessen dringend verbessert werden

--------------------

思路：

*  让找两段文字的公共单词
 *  刚开始觉得输入比较困难，所以没有提前做，后来同学把这道题给A了
 *  我又重新看了这题，也就试了一下
 *  但是因为Dev编译器的原因，因为输入的问题，找了半个小时的bug，最后发现问题出现在编译器身上
 *  果断放弃Dev，转战CodeBlocks
 *  把每一个单词都当成字符串输入，用string  s t r i n g 存每一个单词
 *  开两个string  s t r i n g 数组，然后在进行输入，需要注意的是这是多组输入，还需要注意\#是一段文字输入结束标志
 *  其他的也没什么了，也就是寻找路径的模板问题

--------------------

*  code

//#include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    const int MAXN=120;
    string s;
    string s1[120];
    string s2[120];
    int dp[MAXN][MAXN];
    int vis[MAXN][MAXN];
    
    void dfs(int i,int j)
    {
        if(i==0||j==0)
            return ;
        if(vis[i][j]==1)
        {
            dfs(i-1,j-1);
            cout<<s1[i-1]<<" ";
        }
        else if(vis[i][j]==2)
            dfs(i-1,j);
        else
            dfs(i,j-1);
    
    }
    
    int main()
    {
        while(cin>>s)
        {
            s1[0]=s;
            int len1=1,len2=0;
            while(cin>>s)
            {
                if(s=="#")  break;
                s1[len1++]=s;
            }
            while(cin>>s)
            {
                if(s=="#")  break;
                s2[len2++]=s;
            }
    //      for(int i=1;i<len1;i++)
    //          cout<<s1[i]<<" ";
    //        cout<<endl;
    //      for(int i=0;i<len2;i++)
    //          cout<<s2[i]<<" ";
    //        cout<<endl;
            memset(dp,0,sizeof(dp));
            memset(vis,0,sizeof(vis));
            for(int i=1;i<=len1;i++)
            {
                for(int j=1;j<=len2;j++)
                    if(s1[i-1]==s2[j-1])
                    {
                        dp[i][j]=dp[i-1][j-1]+1;
                        vis[i][j]=1;
                    }
                    else if(dp[i-1][j]>dp[i][j-1])
                    {
                        dp[i][j]=dp[i-1][j];
                        vis[i][j]=2;
                    }
                    else
                    {
                        dp[i][j]=dp[i][j-1];
                        vis[i][j]=3;
                    }
            }
            dfs(len1,len2);
            cout<<endl;
        }
        return 0;
    }