POJ 2250 （LCS，经典输出LCS序列 dfs）

题目链接：

Compromise

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9284		Accepted: 3972		Special Judge

Description

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

Source

Ulm Local 1997

分析：

经典的LCS问题，重点是采用DFS输出此LCS序列中的一个

注意输入两个序列的方式，学习了

注意输入序列的下标从1开始

注意DFS要先递归到最后再输出序列

注意dp的初始化，直接memset就可以

代码如下：

代码如下;

#include<cstring>

#include<cstdio>

#include<string>

#include<iostream>

#include<algorithm>

#define max_v 1005

using namespace std;

string x[max_v],y[max_v];

int dp[max_v][max_v];

int l1,l2;

int dfs(int i,int j)

{

    if(i==||j==)

        return  ;

    if(x[i]==y[j])//来自左上角

    {

        dfs(i-,j-);

        cout<<x[i]<<" ";//先递归到最后再输出，，这样就是顺序的

    }

    else

    {

        if(dp[i-][j]>dp[i][j-])//来自上面

        {

            dfs(i-,j);

        }

        else//来自左边

        {

            dfs(i,j-);

        }

    }

    return ;

}

int main()

{

    string s;

    while(cin>>s)

    {

        l1=l2=;

        if(s!="#")

        {

            x[++l1]=s;

            while(cin>>s&&s!="#")

            {

                x[++l1]=s;

            }

        }

        while(cin>>s&&s!="#")

        {

            y[++l2]=s;

        }

        memset(dp,,sizeof(dp));

        for(int i=; i<=l1; i++)

        {

            for(int j=; j<=l2; j++)

            {

                if(x[i]==y[j])

                {

                    dp[i][j]=dp[i-][j-]+;

                }

                else

                {

                    dp[i][j]=max(dp[i-][j],dp[i][j-]);

                }

            }

        }

        dfs(l1,l2);

        cout<<endl;

    }

    return ;

}

巴特西

POJ 2250 （LCS，经典输出LCS序列 dfs）

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