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
题目大意：给出两篇文章，让你求出两篇文章的公共单词，按顺序输出。
文章？单词？？别想多，单词对于文章，其实和字符对于字符串，”地位上是一样的”（理解一下）。就是找最长公共子序列罢了。
#include<iostream>
#include<vector>
#include<cstring>
#include<string>
using namespace std;
vector<string> s;
vector<string> t;
int dp[1000][1000]; //dp数组
int pre[1000][1000]; //标记路径
vector<string> ans; //保存路径
void output(int x,int y){
if(x==0 || y==0)
return ;
else if(pre[x][y]==1){
output(x-1,y-1);
ans.push_back(s[x-1]);
}
else if(pre[x][y]==2)
output(x-1,y);
else output(x,y-1);
}
int main(){
string r;
int cnt=0;
while(cin>>r){
if(r=="#") cnt++;
if(cnt==0) s.push_back(r);
else if(cnt==1){
if(r=="#"){ //这里要注意一下，不跳过的话t(第二篇文章)开头会多一个"#".
cnt++; continue;
}
t.push_back(r);
}
else if(cnt==2){
int n=s.size(),m=t.size();
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(s[i]==t[j]){
dp[i+1][j+1]=dp[i][j]+1;
pre[i+1][j+1]=1;
}
else{
if(dp[i][j+1]>dp[i+1][j]){
dp[i+1][j+1]=dp[i][j+1];
pre[i+1][j+1]=2;
}
else{
dp[i+1][j+1]=dp[i+1][j];
pre[i+1][j+1]=3;
}
}
}
}
output(s.size(),t.size());
for(int i=0;i<ans.size();i++){
if(i!=ans.size()-1)
cout<<ans[i]<<" ";
else
cout<<ans[i]<<endl;
}
ans.clear(); s.clear(); t.clear(); //不要忘记初始化
cnt=0;
memset(pre,0,sizeof(pre));
memset(dp,0,sizeof(dp));
}
}
return 0;
}