Common Subsequence POJ - 1458（dp）

 Common Subsequence

POJ - 1458



A subsequence of a g[u]iven 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, x
ij = 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.
[/u]
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

思路：慢慢重心开始有贪心转向动态规划了，这题就是简单的动态规划题。以题目的第一组测试数据为例。abcfbc
abfcab。

辅助空间变化示意图



可以看出：

F[i][j]=F[i-1][j-1]+1；（a[i]==b[j]）

F[i][j]=max(F[i-1][j],F[i][j-1])(a[i]!=b[j]);

n由于F(i,j)只和F(i-1,j-1), F(i-1,j)和F(i,j-1)有关, 而在计算F(i,j)时, 只要选择一个合适的顺序, 就可以保证这三项都已经计算出来了, 这样就可以计算出F(i,j). 这样一直推到f(len(a),len(b))就得到所要求的解了.

code：

#include <iostream>
#include <cstring>
using namespace std;
int dp[1005][1005];
string s1,s2;
int main(){
while(cin >> s1 >> s2){
memset(dp,0,sizeof(dp));
int i,j;
for(i = 1; i <= s1.length(); i++){
for(j = 1; j <= s2.length(); j++){
if(s1[i-1] == s2[j-1])
dp[i][j] = dp[i-1][j-1] + 1;
else
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
cout << dp[s1.length()][s2.length()] << endl;
}
return 0;
}