杭电 HDU ACM 1159 Common Subsequence

发布时间：2020-12-13 20:43:40 所属栏目：PHP教程来源：网络整理

导读：Common Subsequence Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26493Accepted Submission(s): 11771 Problem Description A subsequence of a given sequence is the given sequence with som

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26493 Accepted Submission(s): 11771

Problem 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,xij = zj. For example,Z = <a,b,f,c> is a subsequence of X = <a,c,c> with index sequence <1,4,6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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.

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

表示好几天没有a过题目了，缘由好多，平时课程多死，省赛训练也常常跑西区，做的题目也不再是水题了，主要学习进程挺费时间，我相信只要把某个算法学会了，然后

做的类似的题目就不会浪费时间了。今天51 也算有了1点点属于自己的时间，（其实51还是有好多事情要干）。

参考学习了网上这篇博文，和《算法导论》叙述得差不多，

http://blog.csdn.net/yysdsyl/article/details/4226630

代码敲的少啊，连scanf（）语法也卡了好大会儿。不过以后知道了……

其实我感觉　ｓｔｒｉｎｇ处理字符串　也挺方便。

#include<iostream > #include<stdio.h> #include<cstring> const int M=500; using namespace std; int main() { char cnt[M],dic[M]; int c[M][M],m; while(scanf("%s%s",cnt+1,dic+1)!=EOF) { int lencnt=strlen(cnt+1); int lendic=strlen(dic+1); for(int i=1; i<=lencnt; i++) c[i][0]=0; for(int j=0; j<=lendic; j++) c[0][j]=0; for( k=1; k<=lencnt; k++) for( m=1; m<=lendic; m++) { if(cnt[k]==dic[m]) { c[k][m]=c[k⑴][m⑴]+1; } else { c[k][m]=max(c[k⑴][m],c[k][m⑴]); } } cout<<c[lencnt][lendic]<<endl; } return 0; }

（编辑：李大同）

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