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 题意： 这个题目和书上的Lcs 问题差不多 但是他的元素不再是字符 变成了 每个元素都是单词也就是字符串方程：
LCS（i,j）表示数组以数组1的i个为止和以数组2的第j个为止的最长公共子序列，注意是第i个字符为子串的末尾字符（这样才能达到无后效性的目的） 1.LCS（i-1，j-1）+1 （a【i】=b【j】）LCS（i,j）=｛ 2.max｛LCS（i-1，j），LCS（i,j-1）｝ （a【i】！=b【j】）
#include<iostream>
#include<cstring>
#include<string>
#include<vector>
using namespace std;
vector<string> st1,st2;
int dp[110][110],path[110][110];
int tag;
void printpath(int x,int y){
if(x==0||y==0) return;
if(path[x][y]==1){
printpath(x-1, y-1);
if(tag==0){
cout<<st1[x-1];
tag=1;
}
else cout<<" "<<st1[x-1];
}
else if(path[x][y]==2) printpath(x-1, y);
else if(path[x][y]==3)printpath(x, y-1);
}
int main(){
string str;
while(cin>>str){
st1.clear();st2.clear();
st1.push_back(str);
while(cin>>str&&str!="#") st1.push_back(str);
while(cin>>str&&str!="#") st2.push_back(str);
memset(dp,0,sizeof(dp));
memset(path,0,sizeof(path));
int len1=st1.size(),len2=st2.size();
tag=0;
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
if(st1[i-1]==st2[j-1]){
dp[i][j]=dp[i-1][j-1]+1;
path[i][j]=1;
}else if(dp[i-1][j]>dp[i][j-1]){
dp[i][j]=dp[i-1][j];
path[i][j]=2;
}else{
dp[i][j]=dp[i][j-1];
path[i][j]=3;
}
}
}
printpath(len1, len2);
cout<<endl;
}
return 0;
}