UVa 10405 - Longest Common Subsequence,最长公共子序列模板题
2012-08-18 14:13
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【链接】
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=1346
【原题】
Sequence 2:
Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:
is adh of length 3.
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
For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.
【题目大意】
输入两个字符串,求出它们的最长公共子序列有多长
【分析与总结】
最长公共子序列模板题
【代码】
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=1346
【原题】
Problem C: Longest Common Subsequence
Sequence 1:Sequence 2:
Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:
abcdgh aedfhr
is adh of length 3.
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
For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.
Sample input
a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh aedfhrabcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn
Output for the sample input
4 3 26 14
【题目大意】
输入两个字符串,求出它们的最长公共子序列有多长
【分析与总结】
最长公共子序列模板题
【代码】
/* * UVa: 10405 - Longest Common Subsequence * Time: 0.076s * Author: D_Double * */ #include<iostream> #include<cstring> #include<cstdio> using namespace std; char str1[1002],str2[1002]; int d[1002][1002]; int main(){ while(gets(str1)&&gets(str2)){ int len1=strlen(str1), len2=strlen(str2); memset(d, 0, sizeof(d)); for(int i=1; i<=len1; ++i){ for(int j=1; j<=len2; ++j){ if(str1[i-1]==str2[j-1]) d[i][j] = d[i-1][j-1]+1; else d[i][j] = max(d[i-1][j], d[i][j-1]); } } printf("%d\n", d[len1][len2]); } return 0; }
—— 生命的意义,在于赋予它意义。 原创 http://blog.csdn.net/shuangde800 , By D_Double (转载请标明)
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