poj 1458 LCS(最长公共子序列)

Common Subsequence
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 53343 Accepted: 22124
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

dp[i][j]表示 以a的i字符结尾的字符串与以b的j字符结尾的字符串最长的子串长度

如果a[i]==b[j] dp[i][j]=dp[i-1][j-1]+1;
如果a[i]!=b[j] 说明最长子串绝对不是以 a[i]和b[j]结尾的
所以要推到a[i] b[j-1] 和a[i-1]b[j]
dp[i][j]=max(dp[i][j-1],dp[i-1][j])

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define maxn 1010using namespace std;char a[maxn];char b[maxn];int dp[maxn][maxn];int main(){    while(scanf("%s%s",a,b)!=EOF)    {        memset(dp,0,sizeof(dp));        int l1=strlen(a);        int l2=strlen(b);        for(int i=l1;i>0;i--)        {            a[i]=a[i-1];        }        for(int j=l2;j>0;j--)b[j]=b[j-1];        for(int i=1;i<=l1;i++)        {            for(int j=1;j<=l2;j++)            {                if(a[i]==b[j])                {                    dp[i][j]=dp[i-1][j-1]+1;                }                else                {                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);                }            }        }        printf("%d\n",dp[l1][l2]);    }    return 0;}