<span style="color:#000099;">/*
C - 简单dp 例题
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit
Status
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
By Grant Yuan
2014.7.16
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
char a[1000];
char b[1000];
int dp[1000][1000];
int max(int aa,int bb)
{
return aa>=bb?aa:bb;
}
int main()
{ int l1,l2;
while(~scanf("%s%s",a,b)){
l1=strlen(a);
l2=strlen(b);
memset(dp,0,sizeof(dp));
for(int i=0;i<l1;i++)
for(int j=0;j<l2;j++)
{
if(a[i]==b[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]);
}
cout<<dp[l1][l2]<<endl;
}
return 0;
}
</span>原文地址：http://blog.csdn.net/yuanchang_best/article/details/37879679
