poj 1458 Common Subsequence

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#POJ #动态规划 #最长公共子序列

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38710		Accepted: 15576

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 _{i_j} = 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.

对于一个序列来说，他的子序列就是一个有些元素没有的序列。对于一个序列 X = < x1, x2, ..., xm > ,另一个序列 Z = < z1, z2, ..., zk >如果是X的子序列，需要满足，对于X存在一个严格递增的序列< i1, i2, ..., ik > 作为下标，所有的j = 1,2,...,k，x_{i_j} = zj。比如Z = < a, b, f, c >是X = < a, b, c, f, b, c >的子序列。给出两个序列X和Y，找出两者的最长公共子序列。

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.

输入每行是两个字符串，中间包含任意数量的空格。（据说长度不超过200）

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.

输出为输入的两个字符串的最长公共子序列。

Sample Input

abcfbc         abfcab
programming    contest 
abcd           mnp

Sample Output

4
2
0

Source

Southeastern Europe 2003

来自算法导论经典算法。。LCS最长公共子序列。

#include <stdio.h>
#include <string.h>

int max_num(int a, int b);

int main(void){

    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);

    char str_a[222], str_b[222];
    int c[222][222];
    int len_a, len_b;
    int i, j;
    while(scanf("%s%s", str_a + 1, str_b + 1) != EOF){
        printf("%s %s\n", str_a + 1, str_b + 1);
        len_a = strlen(str_a + 1);
        len_b = strlen(str_b + 1);

        for (i = 0;i < len_a + 1;i ++){
            c[i][0] = 0;
        }
        for (j = 0;j < len_b + 1;j ++){
            c[0][j] = 0;
        }

        for (i = 1;i < len_a + 1;i ++)
            for (j = 1;j < len_b + 1;j ++){
                if (str_a[i] == str_b[j]){
                    c[i][j] = c[i - 1][j - 1] + 1;
                }else{
                    c[i][j] = max_num(c[i - 1][j], c[i][j - 1]);
                }
            }
        printf("%d\n", c[len_a][len_b]);
    }
    return 0;
}

int max_num(int a, int b){
    return a > b ? a : b;
}