poj_1458 LCS problem F.最长上升公共子序列

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

 分析：

d[i][j]表示a[1],a[2]……a[i] 和b[1],b[2]……b[i]的最长上升子序列长度.while(a[i]==a[j])  d[i][j]=d[i-1][j-1]+1;   else  d[i][j]=max{d[i-1][j],d[i][j-1]};     时间复杂度为o(n*m).

 

代码及简要分析：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>   //使用cin>>输入需要加上此头文件
using namespace std;
string a,b;
int dp[1000][1000];
int max(int x,int y)
{
    if(x>=y)
        return x;
    else
        return y;
}

int main()
{
    int i,j;
    while(cin>>a)
    {
        cin>>b;
        memset(dp,0,sizeof(dp));//ddp[i][j]为a[0]...a[i]和b[0]...b[j]的最长公共子序列的长度。
        for(i=0;i<a.size();i++)
        {
            for(j=0;j<b.size();j++)
            {
                if(a[i]==b[j])
                    dp[i+1][j+1]=dp[i][j]+1;  //注意此从 dp[1][1]开始
                else
                    dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);

            }
        }
        printf("%d\n",dp[i][j]);
    }
   return 0;
}