poj 1458 Common Subsequence (LCS模版题)

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47946		Accepted: 19749

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

状态转移方程：      

dp[i][j]=dp[i-1][j-1]   (s[i-1]==t[j-1]);

d[i][j]=max(dp[i-1][j], dp[i][j-1]]). (其他)

初始化dp[ ][ ],

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define M 1005
using namespace std;
int dp[M][M];
char str1[1010],str2[1010];
int main()
{

int lena,lenb;
while(~scanf("%s%s",str1,str2))
{
int i,j;
memset(dp,0,sizeof(dp));
lena=strlen(str1);
lenb=strlen(str2);
for(int i=0;i<lenb;i++)
dp[i][0]=0;
for(int j=0;j<lena;j++)
dp[0][j]=0;
for(i=1;i<=lena;i++)
{
for(j=1;j<=lenb;j++)
{
if(str1[i-1]==str2[j-1])
{
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[lena][lenb]);
}
return 0;
}