POJ 2250 Compromise (longest common sub-sequence)

Compromise

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6750		Accepted: 3018		Special Judge

Description

In a few months the European Currency Union would become a reality. However, to join the club, the Maastricht criteria must being fulfilled, and this is not a trivial task for the countries (MA Ybe except for Luxembourg). To enforce that Germany would fulfill the criteria, our government have so many wonderful options (raise taxes, sell stocks, Revalue the Gold reserves,...) That's it's really hard to choose.

Therefore The German government requires a program for the following task:
Both politicians each enter their proposal of "what". The computer then outputs the longest common subsequence of words, the 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 hav E in mind).

Your Country needs This program, so Your job was to write it for us.

Input

The input would contain several test cases.
Each test case consists of texts. Each text was given as a sequence of lower-case words, separated by whitespace, and with no punctuation. Words'll is less than and characters long. Both texts would contain less than words and would 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. If there is more than one such sequence, all one is acceptable. Separate the words by one blank. After the last word, output a newline character.

Sample Input

Die Einkommen der Landwirtesind fuer die abgeordneten ein Buch mit sieben siegelnum dem Abzuhelfenmuessen dringend alle su Bventionsgesetze verbessert werden#die Steuern auf Vermoegen und einkommensollten nach Meinung der Abgeordnetennachdrueck Lich Erhoben Werdendazu muessen die Kontrollbefugnisse der finanzbehoerdendringend Verbessert werden#

Sample Output

Die Einkommen der Abgeordneten Muessen dringend verbessert werden

Idea: Classic DP problem, I mapped a string array into an int array using a map, and then recursively printing the solution

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm>using namespace STD; #include <map>map<string,int> all;char s[50];int s1[110],s2[110];int dp[110][110];string ache[220]; int DP (int x,int y) {if (x<0| |    Y&LT;0) return 0;    int &ans=dp[x][y];    if (dp[x][y]>=0) return ans;    if (S1[x]==s2[y]) ANS=DP (x-1,y-1) +1;    else return Ans=max (DP (X-1,Y), DP (X,Y-1)); return ans;} void print (int x,int y) {if (x<0| |    Y&LT;0) return;        if (S1[x]==s2[y]) {print (x-1,y-1);        if (DP (x, y) ==1) cout<<ache[s1[x]];    else cout<< ' <<ache[s1[x]];        } else {if (DP (x,y-1) &GT;DP (x-1,y)) print (x,y-1);    else print (x-1,y);    }}void Ini () {memset (dp,-1,sizeof (DP)); All.clear ();}        int main () {while (~SCANF ("%s", s)) {ini ();        if (s[0]== ' # ') continue;        int top=0;        int l1=0,l2=0;        String t=s;  if (All.count (t) ==0) all[t]=++top,ache[top]=t;      S1[L1++]=ALL[T];            while (scanf ("%s", s)) {if (s[0]== ' # ') break;            T=s;            if (All.count (t) ==0) all[t]=++top,ache[top]=t;        S1[L1++]=ALL[T];            } while (scanf ("%s", s)) {if (s[0]== ' # ') break;            T=s;            if (All.count (t) ==0) all[t]=++top,ache[top]=t;        S2[L2++]=ALL[T];        } dp[0][0]= (S1[0]==s2[0]);        DP (l1-1,l2-1);        Print (l1-1,l2-1);    Puts (""); } return 0;}

POJ 2250 Compromise (longest common sub-sequence)