HDU-1159 Common Subsequence(动态规划2）

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

x的一个子序列相应于下标序列{1，2,3.....m}的一个子序列，故X共有2^m个子序列，直接爆，需要指数时间。为此，考虑能否用动态规划方法求解。
先找出它是否满足最优子结构性质：
设序列X={x1,...xm}和y={y2,....yn},子序列为z={z1,...zk}
(1)如果 xm=yn，则zk=xn=yn；且z（k-1）是x（m-1）和y（n-1）的最长公共子序列。
   找出x（m-1）和y（n-1）的最长公共子序列，然后在其尾部加上xm（=yn）即可得x和y的一个最长公共    子序列
（2）如果xm！=yn，且zk！=xm 则z 是x（m-1）和y的最长公共子序列。
     如果xm！=yn，且zk！=yn 则z 是x  和y（n-1的最长公共子序列。
    找出x（m-1）和y的一个最长公共子序列及x和y（n-1）的最长公共子序列。这两个公共子序列中较长者即为X和Y的一个最长公共子序列。
   如 1 2 3 4 7
      1 2 4 5    
   或 1 2 3 4 9
      1 2 3 4 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    char a[1001],b[1001];
    int n,m,i,j,c[1001][1001];
    while(scanf("%s%s",&a,&b)!=EOF)
    {

        n=strlen(a);
        m=strlen(b);
        if(n==0||b==0)
        {
            printf("0
");
            break;
        }
        memset(c,0,sizeof(c));
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                if(a[i-1]==b[j-1])
                {
                    c[i][j]=c[i-1][j-1]+1;
                }
                else
                {
                    c[i][j]=max(c[i-1][j],c[i][j-1]);
                }
            }
        }
        printf("%d
",c[n][m]);
    }
    return 0;
}