poj1458 Common Subsequence

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 57155 Accepted: 23876

Description

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.

Input

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.

Output

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.

Sample Input

abcfbc abfcab

programming contest

abcd mnp

Sample Output

4

2

0

Source

Southeastern Europe 2003

这个也是用动态规划来做才能ac的题目。比较倒霉的是，我当时因为没有判断结束输入导致超时。

说一下这题的思路

最关键的一点就是确定状态以及状态转移方程

设置两个变量i,j分别表示在两个字符串s1,s2中所在的位置。

1.if(i == 0 || i == j) ,maxLen[i][j] = 0;

2.if(s1[i] == s2[j]) ,maxLen[i][j] = maxLen[i-1][j-1] + 1;

3.if(s1[i] != s2[j]),maxLen[i][j] = max(maxLen[i-1][j],maxLen[i][j-1]);

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
char s1[1000];
char s2[1000];
int maxLen[1000][1000];
int main(){
while(cin >> s1 >> s2){
int len1, len2;
len1 = strlen(s1);
len2 = strlen(s2);

for(int i = 0;i <max(len1,len2); ++i)
{
maxLen[i][0] = 0;
maxLen[0][i] = 0;
}
for(int i = 1;i <= len1; ++i)
for(int j = 1;j <= len2; ++j){
if(s1[i-1] == s2[j-1]) {
maxLen[i][j] = maxLen[i-1][j-1] + 1;
continue;
}
maxLen[i][j] = max(maxLen[i-1][j],maxLen[i][j-1]);
}
cout << maxLen[len1][len2] << endl;
}
return 0;
}

中国大学mooc 程序设计与算法(二) 北京大学 郭炜