Problem Links:
uva00531, zoj1939,Problem:
Compromise
| Compromise |
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 Specification
The input file 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 Specification
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 Code:
import java.io.BufferedReader;import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
/**
* Problem: 531, Compromise.
* Using Longest Common Subsequence.
*/
/**
* @author antonio081014
* @since Jan 29, 2012, 4:55:25 PM
*/
class Main {
public static List<String> listA;
public static List<String> listB;
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String strLine;
for (int t = 0;; t++) {
strLine = br.readLine();
if (strLine == null || strLine.length() == 0
|| strLine.compareTo("#") == 0)
break;
listA = new ArrayList<String>();
listB = new ArrayList<String>();
while (strLine.compareTo("#") != 0) {
String[] strs = strLine.split("\\s");
for (int i = 0; i < strs.length; i++)
listB.add(strs[i]);
strLine = br.readLine();
}
do {
strLine = br.readLine();
String[] strs = strLine.split("\\s");
for (int i = 0; i < strs.length; i++)
listA.add(strs[i]);
} while (strLine.compareTo("#") != 0);
print(LCS());
}
}
public static int[][] LCS() {
int N = listA.size();
int M = listB.size();
int[][] dp = new int[N + 1][M + 1];
int[][] parent = new int[N + 1][M + 1];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++) {
if (listA.get(i - 1).compareTo(listB.get(j - 1)) == 0) {
dp[i][j] = 1 + dp[i - 1][j - 1];
parent[i][j] = 1;
}
else {
dp[i][j] = dp[i - 1][j];
parent[i][j] = 2;
if (dp[i][j - 1] > dp[i - 1][j]) {
dp[i][j] = dp[i][j - 1];
parent[i][j] = 3;
}
}
}
}
return parent;
}
public static void print(int[][] parent) {
List<String> list = new ArrayList<String>();
for (int i = listA.size(), j = listB.size(); i > 0 && j > 0;) {
if (parent[i][j] == 1) {
list.add(listA.get(i - 1));
i--;
j--;
}
else {
if (parent[i][j] == 2)
i--;
else if (parent[i][j] == 3)
j--;
}
}
for (int i = list.size() - 1; i > 0; i--)
System.out.print(list.get(i) + " ");
System.out.println(list.get(0));
}
}
as 
where 
denotes the ranking of event i in the correct chronological order and a sequence of student responses 
where 
denotes the chronological rank given by the student to event i; determine the length of the longest (not necessarily contiguous) sequence of events in the student responses that are in the correct chronological order relative to each other.
. The second line will contain nintegers, indicating the correct chronological order of n events. The remaining lines will each consist of n integers with each line representing a student's chronological ordering of the n events. All lines will contain n numbers in the range 
, with each number appearing exactly once per line, and with each number separated from other numbers on the same line by one or more spaces.
(length, width, and height). In 6 dimensions it is, perhaps, unclear what the box (4,5,6,7,8,9) represents; but we can analyze properties of the box such as the sum of its dimensions.
such that each box 
nests in box 
( 
.
) nests in a box E = ( 
) if there is some rearrangement of the 
such that when rearranged each dimension is less than the corresponding dimension in box E. This loosely corresponds to turning box D to see if it will fit in box E. However, since any rearrangement suffices, box D can be contorted, not just turned (see examples below).
of
such that ( 
) ``fits'' in ( 
for all 
.
line in the sequence ( 
) gives the measurements for the 