POJ_1458_Common Subsequence

Common Subsequence

Time Limit: 1000MS
Memory Limit: 10000KTotal Submissions: 54853
Accepted: 22866

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         abfcabprogramming    contest abcd           mnp

Sample Output

420

Source

Southeastern Europe 2003

• LCS就不多说了，看代码

#include <iostream>#include <string>#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>using namespace std;typedef long long           LL ;typedef unsigned long long ULL ;const int    maxn = 1e3 + 10   ;const int    inf  = 0x3f3f3f3f ;const int    npos = -1         ;const int    mod  = 1e9 + 7    ;const int    mxx  = 100 + 5    ;const double eps  = 1e-6       ;const double PI   = acos(-1.0) ;char a[maxn], b[maxn];int la, lb, dp[maxn][maxn];int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(~scanf("%s",a+1)){scanf("%s",b+1);la=strlen(a+1);lb=strlen(b+1);memset(dp,0,sizeof(dp));for(int i=1;i<=la;i++)for(int j=1;j<=lb;j++)if(a[i]==b[j])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);printf("%d\n",dp[la][lb]);}return 0;}