[POJ] 1458 Common Subsequence
Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 56614 Accepted: 23609 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. 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 Source Southeastern Europe 2003
LCS问题
f[i][j]表示S长度为i的前缀和T长度为j的前缀的LCS
f[i][j]=f[i-1][j-1]+1 (s[i]==t[j])
f[i][j]=max( f[i-1][j] , f[i][j-1] ) (else)
//Writer:GhostCai && His Yellow Duck #include<iostream> #include<cstring> #define MAXN 500 using namespace std; char s[MAXN],t[MAXN]; int lens,lent; int f[MAXN][MAXN]; int main(){ while(cin>>s+1>>t+1){ memset(f,0,sizeof(f)); lens=strlen(s+1); lent=strlen(t+1); for(int i=1;i<=lens;i++){ for(int j=1;j<=lent;j++){ if(s[i]==t[j]){ f[i][j]=f[i-1][j-1]+1; }else{ f[i][j]=max(f[i-1][j],f[i][j-1]); } } } cout<<f[lens][lent]<<endl; } return 0; }
本文来自博客园,作者:GhostCai,转载请注明原文链接:https://www.cnblogs.com/ghostcai/p/9247508.html
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