題目鏈接：

http://poj.org/problem?id=2250

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： 經(jīng)典的LCS問(wèn)題，重點(diǎn)是采用DFS輸出此LCS序列中的一個(gè) 注意輸入兩個(gè)序列的方式，學(xué)習(xí)了 注意輸入序列的下標(biāo)從1開(kāi)始 注意DFS要先遞歸到最后再輸出序列 注意dp的初始化，直接memset就可以 代碼如下： 代碼如下; #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==0||j==0)return 0 ;if(x[i]==y[j])//來(lái)自左上角 {dfs(i-1,j-1);cout<<x[i]<<" ";//先遞歸到最后再輸出，，這樣就是順序的 }else{if(dp[i-1][j]>dp[i][j-1])//來(lái)自上面 {dfs(i-1,j);}else//來(lái)自左邊 {dfs(i,j-1);}}return 0; } int main() {string s;while(cin>>s){l1=l2=0;if(s!="#"){x[++l1]=s;while(cin>>s&&s!="#"){x[++l1]=s;}}while(cin>>s&&s!="#"){y[++l2]=s;}memset(dp,0,sizeof(dp));for(int i=1; i<=l1; i++){for(int j=1; j<=l2; j++){if(x[i]==y[j]){dp[i][j]=dp[i-1][j-1]+1;}else{dp[i][j]=max(dp[i-1][j],dp[i][j-1]);}}}dfs(l1,l2);cout<<endl;}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/yinbiao/p/9067766.html

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