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  • hdu 1159(最长公共子序列)

    入门题。。。

    Common Subsequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 15396    Accepted Submission(s): 6406


    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
     
    Recommend
    Ignatius
     
     
    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    using namespace std;
    
    #define N 1010
    
    char s[N],t[N];
    int dp[N][N];
    
    int main()
    {
        while(scanf("%s%s",s,t)!=EOF)
        {
            int n=strlen(s);
            int m=strlen(t);
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++)
                for(int j=1;j<=m;j++)
                {
                    if(s[i-1]==t[j-1]) dp[i][j]=dp[i-1][j-1]+1;
                    else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            printf("%d\n",dp[n][m]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chenhuan001/p/2984432.html
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