HDU 1159 Common Subsequence 最长公共子序列

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15540    Accepted Submission(s): 6469

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 abfcabprogramming contest abcd mnp

 

Sample Output

420

 

Source

Southeastern Europe 2003

 

Recommend

Ignatius

动态规划中典型的LCS问题。只要把握如下核心：

if(s1[i]==s2[j])

len[i][j]=len[i-1][j-1]+1

else

len[i][j]=max(len[i-1][j],len[i][j-1])

用一个矩阵表示，举个例子，比如programn contest

         p   r   o    g  r   a   m   n

        0    0   0   0  0   0   0    0

c  0  0   0   0    0  0   0  0   0

o  0  0   0   1    1  1  1   1    1 

n  0  0   0   1    1  1   1   1    2

t   0   0   0   1    1  1   1   1     2

e  0   0   0   1    1  1   1    1    2

s  0   0    0   1   1   1  1    1    2

t   0    0    0   1   1   1  1   1     2

最后直接输出len[MAX_LEN][MAX_LEN]

#include<stdio.h>#include<string.h>char s1[1007],s2[1007];int dp[1007][1007];int max(int x,int y){    return x>y?x:y;}int main(){    while(scanf("%s%s",s1+1,s2+1)!=EOF)    {        int len1,len2,i,j;        len1=strlen(s1+1);        len2=strlen(s2+1);        for(i=0;i<=len1;i++)        dp[i][0]=0;        for(j=0;j<=len2;j++)        dp[0][j]=0;        for(i=1;i<=len1;i++)        for(j=1;j<=len2;j++)        {            if(s1[i]==s2[j])            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[len1][len2]);    }    return 0;}