总时间限制:
1000ms
内存限制:
65536kB
描述
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, …, xm > another sequence Z = < z1, z2, …, zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, …, ik > of indices of X such that for all j = 1,2,…,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
输入
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
输出
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
样例输入
abcfbc abfcab
programming contest
abcd mnp
样例输出
4
2
0
直接上代码
#include <iostream>
#include<string>
#include <algorithm>
using namespace std;
int main()
{
string str1,str2;
while(cin>>str1>>str2)
{
int dp[str1.length()][str2.length()];
for (int i = 0; i < str1.length(); ++i) {
for (int j = 0; j < str2.length(); ++j) {
dp[i][j]=0;
}
}
int exam = 0;
for (int i = 0; i < str1.length(); ++i) {
if(str2[0]==str1[i]) exam=1;
dp[i][0]=exam;
}
exam = 0;
for (int i = 0; i < str2.length(); ++i) {
if(str1[0]==str2[i]) exam = 1;
dp[0][i]=exam;
}
for (int i = 1; i < str1.length(); ++i) {
for (int j = 1; j < str2.length(); ++j) {
if(str1[i]==str2[j]){
dp[i][j]=dp[i-1][j-1]+1;
}
else{
dp[i][j]=max(dp[i][j-1], dp[i-1][j]);
}
}
}
exam = 0;
for (int i = 0; i < str1.length(); ++i) {
for (int j = 0; j < str2.length(); ++j) {
exam=max(exam, dp[i][j]);
}
}
cout<<exam<<endl;
}
return 0;
}
java
import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner s = new Scanner(System.in);
String str1, str2;
while (s.hasNext()) {
str1 = s.next();
str2 = s.next();
char[] str1_c = str1.toCharArray();
char[] str2_c = str2.toCharArray();
ArrayList<ArrayList> dp = new ArrayList<ArrayList>();
for (int i = 0; i < str1_c.length; i++) {
ArrayList n_e = new ArrayList();
for (int j = 0; j < str2_c.length; j++) {
n_e.add(0);
}
dp.add(n_e);
}
int exam = 0;
for (int i = 0; i < str1_c.length; i++) {
if (str2.charAt(0) == str1.charAt(i)) exam = 1;
dp.get(i).set(0, exam);
}
exam = 0;
for (int i = 0; i < str2_c.length; i++) {
if (str1.charAt(0) == str2.charAt(i)) exam = 1;
dp.get(0).set(i, exam);
}
for (int i = 1; i < str1_c.length; i++) {
for (int j = 1; j < str2_c.length; j++) {
if (str1.charAt(i) == str2.charAt(j)) {
//相等
dp.get(i).set(j, (int) dp.get(i - 1).get(j - 1) + 1);
} else {
//不相等
int maxs = Math.max((int) dp.get(i).get(j - 1), (int) dp.get(i - 1).get(j));
dp.get(i).set(j, maxs);
}
}
}
exam = 0;
for (int i = 0; i < str1_c.length; i++) {
for (int j = 0; j < str2_c.length; j++) {
if (exam < (int) dp.get(i).get(j)) {
exam = (int) dp.get(i).get(j);
}
}
}
System.out.println(exam);
}
}
}
/*
7
1 7 3 5 9 4 8
*/