Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 29009    Accepted Submission(s): 12992

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, b, c, f, b, c> with index sequence <1, 2, 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

 

Source

Southeastern Europe 2003

 

Recommend

Ignatius   |   We have carefully selected several similar problems for you:  1069 2084 1203 1160 2602

题意：求两个字符串的最大公共子序列长度，可以不连续。

思路：dp，用过string函数中的strchr（a，ch）可惜会wa

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;char a[1005];char b[1005];int dp[1005][1005];int max(int a,int b){    return a>b?a:b;}int main(){    while(~scanf("%s%s",a,b))    {        memset(dp,0,sizeof(dp));        int len1=strlen(a);        int len2=strlen(b);        for(int i=1;i<=len1;i++)        {            for(int j=1;j<=len2;j++)            {                if(a[i-1]==b[j-1])                {                    dp[i][j]=dp[i-1][j-1]+1;                }                else                {                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);                }            }        }        printf("%d\n",dp[len1][len2]);    }    return 0;}
         
        
       
      
     

用过string函数中的strchr（a，ch）可惜会wa，数据能过

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;char a[100005];char b[100005];char c;int visit[105];int main(){ int t,len; while(~scanf("%s%s",a,b)) { t=0; len=strlen(b); memset(visit,0,sizeof(visit)); for(int i=0; i