UVA10405 Longest Common Subsequence【LCS+DP】

标签：a10   put   ems   注意   模板   pairs   ==   efi   mos   

Given two sequences of characters, print the length of the longest common subsequence of both sequences.
????Sequence 1:

????Sequence 2:

????For example, the longest common subsequence of the following two sequences ‘abcdgh’ ans ‘aedfhr’ is ‘adh’ of length 3.

Input
Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

Output
For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample Input
bcacbcabbaccbab
bccabccbbabacbc
a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Sample Output
11
4
3
26
14

问题链接：UVA10405 Longest Common Subsequence
问题描述：（略）
问题分析：
????动态规划问题，是一个标准模板题，套模板就可以了。
????需要注意字符串长度！
程序说明：
????需要注意，要用行读入函数gets()读入数据，不能使用格式化输入"%s%s"，不然会WA。
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* UVA10405 Longest Common Subsequence */

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

#define MAX(x,y) (((x) > (y)) ? (x) : (y))

#define N 1000
char a[N + 2], b[N + 2];
int dp[N + 1][N + 1];

int lcs(char *p, char *q)
{
    memset(dp, 0, sizeof(dp));

    int len1 = strlen(p + 1);
    int len2 = strlen(q + 1);
    int i, j;
    for(i = 1; i <= len1; i++)
        for(j = 1; j <= len2; j++) {
            if(p[i] == q[j])
                dp[i][j] = dp[i - 1][j - 1] + 1;
            else
                dp[i][j] = MAX(dp[i - 1][j], dp[i][j - 1]);
        }

    return dp[len1][len2];
}

int main(void)
{
    while(gets(a + 1), gets(b + 1))
        printf("%d\n", lcs(a, b));

    return 0;
}

UVA10405 Longest Common Subsequence【LCS+DP】

标签：a10   put   ems   注意   模板   pairs   ==   efi   mos   

原文地址：https://www.cnblogs.com/tigerisland45/p/10054428.html