UVa 10405 - Longest Common Subsequence，最长公共子序列模板题

【链接】

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
aedfhr
abcdefghijklmnopqrstuvwxyz
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)