[NWPU][2014][TRN][6]动态规划第一讲——简单线性dp_C题

题目：

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

题解：

确定状态：f[i][j]表示前一个字符串的前i位与后一个字符串的前j位的最长公共子序列长度；

状态转移方程： f[i][j]=f[i-1][j-1]+1   x[i]==y[j]； f[i][j]=max(f[i - 1][j] ,f[i][j - 1])    x[i]!=y[j]； 注意一下边界，这是做DP肯定要考虑的；
代码：

#include<stdio.h>
#include<string.h>
int f[1003][1003];
int maxint(int a,int b)
{
return a>b?a:b;
}
int main()
{
char a[1003],b[1003];
int i,j,len1,len2;
while(~scanf("%s %s",a,b))
{
len1=strlen(a);
len2=strlen(b);
for(i=0;i<len1;i++)
f[0][i]=0;
for(j=0;j<len2;j++)
f[j][0]=0;
for(i=1;i<=len1;i++)
for(j=1;j<=len2;j++)
if(a[i-1]==b[j-1])
f[i][j]=f[i-1][j-1]+1;
else f[i][j]=maxint(f[i-1][j],f[i][j-1]);
printf("%d\n",f[len1][len2]);
}
return 0;
}