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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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.
Input
abcfbc abfcab programming contest abcd mnp
Output
4 2 0
题意很好理解就看两个字符串公共的长度是多少，然后计算完毕输出就可以了。刚开始我没用的dp做，简单的字符串处理结果错了，后来用dp就过了。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
char s[1010],t[1010];
int dp[1010][1010];
int main()
{
while(~scanf("%s%s",s,t))
{
memset(dp,0,sizeof(dp));
n = strlen(s);
m = strlen(t);
for( int i = 0; i < n; i++ )
{
for( int j = 0; j < m; j++ )
{
if( s[i] == t[j] )
dp[i+1][j+1] = dp[i][j] + 1;
else
dp[i+1][j+1] = max(dp[i][j+1],dp[i+1][j]);
}
}
printf("%d\n",dp[n][m]);
/*int sum=0,flag=0;失败的字符串简单处理
for(int i=0; i<m; i++)
{
for(int j=flag; j<n; j++)
{
if(t[i]==s[j])
{
sum++;
flag=j;
}
}
}
printf("%d\n",sum);
*/
}
return 0;
}