LCS 线性DP入门

C - Common Subsequence

 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. 

Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0
题意：最长上升子序列
思路：基础线性DP
状态：DP【i】【j】表示截至字符串s1前i个元素，字符串s2前j个元素最长上升子序列
目标：MAX（DP【i】【j】）
状态转移：if(s1[i]==s2[k])     dp[i][k]=max(dp[i][k],dp[i-1][k-1])+1;
               else                      dp[i][k]=max(dp[i-1][k],dp[i][k-1]);
决策：s1【i】与s2【j】是否相等
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#define ll long long int
using namespace std;
int dp[1005][1005];
char s1[1005],s2[1005];
int main()
{
    while(scanf("%s%s",s1+1,s2+1)!=EOF)
    {
        memset(dp,0,sizeof(dp));//初始化 
        int ss1=strlen(s1+1),ss2=strlen(s2+1);
        int MAX=-1;
        for(int i=1;i<=ss1;i++)
           for(int k=1;k<=ss2;k++)
           {
                   if(s1[i]==s2[k]) 
                      dp[i][k]=max(dp[i][k],dp[i-1][k-1])+1;
                   else
                      dp[i][k]=max(dp[i-1][k],dp[i][k-1]);
                   MAX=max(MAX,dp[i][k]);
           }
        cout<<MAX<<endl;
       }
    
    return 0;
}