UVA 10405 (13.07.26)

最新推荐文章于 2014-07-09 16:49:28 发布

原创最新推荐文章于 2014-07-09 16:49:28 发布 · 737 阅读

0 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种使用动态规划解决最长公共子序列问题的方法。通过一个具体的C语言实现示例，详细展示了如何构造状态转移方程，并利用二维数组来记录最长公共子序列的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem C: Longest Common Subsequence

Sequence 1:

Sequence 2:

Given two sequences of characters, print thelength of the longest common subsequence of both sequences. Forexample, the longest common subsequence of the following twosequences:

abcdgh
aedfhr

is adh of length 3.

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

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

Sample input

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

求最长公共子序列, 动规的经典题目~
DP[i][j]数组的意思为: 以X数组i位置的字符结尾的, 在Y数组j位置时, 最长的序列长度~
找出状态转移方程说难也难, 说容易也容易, 找出来就发现问题简单了~

必须AC:
#include<stdio.h>
#include<string.h>

char str1[1005];
char str2[1005];
int DP[1005][1005];

int main() {
	while(gets(str1) != NULL) {

		gets(str2);
		int len1, len2;
		memset(DP, 0, sizeof(DP));

		len1 = strlen(str1);
		len2 = strlen(str2);

		for(int i = 0; i < len1; i++) {
			for(int j = 0; j < len2; j++) {
				if(str1[i] == str2[j])
					DP[i+1][j+1] = DP[i][j] + 1;
				else {
					if(DP[i][j+1] >= DP[i+1][j])
						DP[i+1][j+1] = DP[i][j+1];
					else
						DP[i+1][j+1] = DP[i+1][j];
				}
			}
		}
		printf("%d\n", DP[len1][len2]);
	}
}