UVA 531Compromise

   Compromise 
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 Specification 
The input file 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 Specification 
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

/*最长公共子序列+途径输出Sample Input die einkommen der landwirtesind fuer die abgeordneten ein buch mit sieben siegelnum dem abzuhelfenmuessen dringend alle subventionsgesetze verbessert werden#die steuern auf vermoegen und einkommensollten nach meinung der abgeordnetennachdruecklich erhoben werdendazu muessen die kontrollbefugnisse der finanzbehoerdendringend verbessert werden#Sample Output die einkommen der abgeordneten muessen dringend verbessert werden题意：输入两个字符串，每个字符串遇到 # 结束，判断两个字符串的最长公共子序列，输出最长公共串，题目要求最长公共串至少有一个。本题的难点在于路径的记录，想了好久，有些细节问题还是没有注意，看了看大牛的博客，用两个数组标记上一步的位置，全部访问完后再往前遍历，用栈存储或者用递归都可以，本题用的栈。*/#include <iostream>#include <algorithm>#include <stdio.h>#include <string>#include <string.h>#include <stack>using namespace std;int dp[110][110];int pos_a[110][110];int pos_b[110][110];string a[110],b[110];int main(){    int i,j;    while(cin >> a[1])    {        ///输入两个字符串        for(i = 2; ;i ++)        {            cin >> a[i];            if(a[i] == "#")            {                i --;                break;            }        }        for(j = 1; ;j ++)        {            cin >> b[j];            if(b[j] == "#")            {                j --;                break;            }        }        memset(dp,0,sizeof(dp));        memset(pos_a,0,sizeof(pos_a));        memset(pos_b,0,sizeof(pos_b));        int n = i ,m = j;        for(i = 1; i <= n; i ++)            for(j = 1; j <= m; j ++)            {                ///每种情况的上一步都记录下来                if(a[i] == b[j])                {                    dp[i][j] = dp[i-1][j-1] + 1;                    pos_a[i][j] = i-1;                    pos_b[i][j] = j-1;                }                else                {                    if(dp[i-1][j] > dp[i][j-1])                    {                        dp[i][j] = dp[i-1][j];                        pos_a[i][j] = i-1;                        pos_b[i][j] = j;                    }                    else                    {                        dp[i][j] = dp[i][j-1];                        pos_a[i][j] = i;                        pos_b[i][j] = j-1;                    }                }            }        stack<string>s;        while(n >= 1 & m >= 1)        {            if(a[n] == b[m])///如果相等的话，入栈                s.push(a[n]);            ///注意这个赋值，不然到下面会产生错误            ///访问上一个位置            int t = n;            n = pos_a[n][m];            m = pos_b[t][m];        }        ///输出        while(s.size() > 1)        {            cout << s.top() << " ";            s.pop();        }        cout << s.top()<<endl;    }    return 0;}