动态规划3之1002

1 题目编号：1002

2 题目内容：

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. <br>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. <br>

 

Sample Input

abcfbc abfcab
programming contest
abcd mnp

 

Sample Output

4
2
0

 

3 题意：求出两个串的公共子序列的长度。

4 解题思路形成过程：p[i][j]表示s1长度为i的前缀和s2长度为j的前缀的最长公共子序列的长度。考虑s1的第i+1个字符位置要么匹配。要么不匹配。匹配的话只能和前j个位置匹配。如果s1[i+1]==s2[j]的话那么dp[i+1][j]=dp[i][j-1]+1很显然成立。如果s1[i+1]!=s2[j]的话i+1只能和前s2 的前j-1个位置匹配。dp[i+1][j]=dp[i+1][j-1]不匹配前s1的i+1个位置的话。dp[i+1][j]=dp[i][j]。那么dp[i+1][j]=max(dp[i][j],dp[i+1][j-1])这样就把s1的第i+1个位置确定了。
5 代码：#include <iostream>  

#include<stdio.h>  

#include<string.h>  

using namespace std;  

const int maxn=1010;  

int dp[maxn][maxn];  

char s1[maxn],s2[maxn];  

int main()  

{  

    int i,j,len1,len2;  

  

    while(~scanf("%s%s",s1+1,s2+1))  

    {  

        len1=strlen(s1+1);  

        len2=strlen(s2+1);  

        memset(dp[0],0,sizeof dp[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;  

}