POJ 2250 Compromise(文章的最长公共子序列LCS)

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:

       给你两段由空格分隔的语句, 要你求该两段语句的最长公共子序列. 且随便输出一个解即可. 注意每个单词需要看成我们一般处理字符串子序列的一个单独字符. 即每个单词是一个整体.

思路：

       与往常计算最长公共子序列一样的方式即可，利用string数组（等效于二维字符数组） 然后用DFS输出序列即可.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdio>
typedef long long ll;
#define N 105
using namespace std;

string s1
;
string s2
;
int n,m;
int dp

;

void dfs(int i,int j)
{
if(i==0 || j==0) return ;
else if(s1[i]==s2[j])
{
dfs(i-1,j-1);
cout<<s1[i]<<" ";
}
else
{
if(dp[i-1][j]>dp[i][j-1])
dfs(i-1,j);
else dfs(i,j-1);
}
}

int main()
{
string s;
while(cin>>s)
{
n=m=0;
if(s!="#")
{
s1[++n]=s;
while(cin>>s && s!="#")
{
s1[++n]=s;
}
}
while(cin>>s && s!="#")
{
s2[++m]=s;
}

//递推过程
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(s1[i]==s2[j])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j], dp[i][j-1]);
}

//输出结果
dfs(n,m);
cout<<endl;
}
return 0;
}