题解 | #Common Subsequence#

原题链接

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, x_ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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.

最长公共子序列：DP

思路：s1[i]与s2[j]是否匹配，dp[i][j]:以s[i]与s[j]为末尾匹配的最长公共序列长度

• ①是：dp[i][j]=dp[i-1][j-1]+1 都往前一个字符，再匹配
• ②否：dp[i][j]=max(dp[i-1][j],dp[i][j-1]) s1往前一个字符，再匹配 或者 s2往前一个字符，再匹配
• 注意s1[0]、s2[0]为哨兵，表示此时匹配长度为0，不进行实际匹配

#include<iostream>
#include<string>
using namespace std;

const int maxn=5001;

string s1,s2;
int dp[maxn][maxn];

void longest(int m1,int m2){
    for(int i=0;i<m1;i++){
        for(int j=0;j<m2;j++){
            if(i==0||j==0)dp[i][j]=0;//有一个字符串匹配位置为0
            else 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]);
            }
        }
    }
}

int main(){
    while(cin>>s1>>s2){
        s1="0"+s1;
        s2="0"+s2;
        int m1=s1.size();
        int m2=s2.size();
        longest(m1,m2);
        printf("%d\n",dp[m1-1][m2-1]);
    }
    return 0;
}