【HDU】1159 Common Subsequence （DP、最長公共子序列）

題目內容

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

題目大意

求最長公共子序列

解題思路

DP，dp[i][j]表示a[0]-a[i]和b[0]-b[j]間的最長公共子序列。
打表可知：當a[i]==b[j]時，dp[i][j]=dp[i-1][j-1]+1;
當a[i]!=b[j]時，dp[i][j]=max(dp[i-1][j],dp[i][j-1]).

AC程式碼

#include<iostream>
 

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<math.h>
#include<limits.h>
#include<stack>
#include<queue>
#define LL long long
using namespace std;
string a,b;
int dp[1002][1002];
int main()
{
    int
 
 lena,lenb,i,j;
    while(cin>>a)
    {
        cin>>b;
        lena=a.length();
        lenb=b.length();
        memset(dp,0,sizeof(dp));
        for(i=0;i<lena;i++)
        {
            for(j=0;j<lenb;j++)
            {
                if(a[i]==b[j])
                    dp[i+1][j+1]=dp[i][j]+1;
                else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
            }
        }
        printf("%d\n",dp[lena][lenb]);
    }
    return 0;
}