（线性dp，LCS） POJ 1458 Common Subsequence

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 65333		Accepted: 27331

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, 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.

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

最长公共子序列问题（LCS） 其状态转换式为：A[i] = A[j]时,d(i,j) = d(i-1,j-1) + 1,否则d(i,j) = max{d(i-1,j),d(i,j-1)}
这个用char数组吧，用string可能出错，。。。打表
C++代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 10000;
int dp[maxn][maxn];
char s1[maxn];
char s2[maxn];
int len1,len2;
int main(){
    while(~scanf("%s%s",s1,s2)){
        len1 = strlen(s1);
        len2 = strlen(s2);
        for(int i = 0; i <= len1; i++){
            dp[i][0] = 0;
        }
        for(int j = 0; j <= len2; j++){
            dp[0][j] = 0;
        }
        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                if(s1[i-1] == s2[j-1])
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = max(dp[i][j-1],dp[i-1][j]);
            }
        }
        printf("%d\n",dp[len1][len2]);
    }
    return 0;
}