kuangbin求带飞DP1 Common Subsequence(LCS)

    xiaoxiao2025-02-02  11

    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. 

    Sample Input

    abcfbc abfcab programming contest abcd mnp

    Sample Output

    4 2 0 // // main.cpp // Common Subsequence // // Created by 张嘉韬 on 16/8/12. // Copyright © 2016年 张嘉韬. All rights reserved. // #include <iostream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; const int maxn=10000; int f[maxn][maxn]; int main(int argc, const char * argv[]) { //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin); string a,b; while(cin>>a) { int r=0; cin>>b; int lena=a.length(); int lenb=b.length(); for(int i=0;i<=lena+1;i++) { for(int j=0;j<=lenb+1;j++) { f[i][j]=0; } } for(int i=0;i<lena;i++) { for(int j=0;j<lenb;j++) { //cout<<"("<<a[i]<<b[i]<<")"; if(a[i]==b[j]) {f[i+1][j+1]=f[i][j]+1;} else {f[i+1][j+1]=max(f[i][j+1],f[i+1][j]);} if(f[i+1][j+1]>r) r=f[i+1][j+1]; //cout<<f[i+1][j+1]<<" "; } //cout<<endl; } printf("%d\n",r); } return 0; }
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