POJ 1458 Common Subsequence（DP）

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本文解析了最长公共子序列(LCS)问题的算法实现，通过动态规划方法找到两个序列的最长公共子序列长度，并提供了完整的C++代码实现。

题目链接：http://poj.org/problem?id=1458

题目大意：给定a，b两个序列，求最长公共子序列

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 _{i_j} = 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

思路：

根据LCS问题的最优子结构性质，可得状态转移方程：

C [i, j] = ⎧ ⎩ ⎨ 0 ， C [i - 1, j - 1] + 1 ， M A X (C [i, j - 1], C [i - 1, j]) 当 i = 0 或 j = 0 当 i, j > 0 且 x i = y j 当 i, j > 0 且 x i \neq y j

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int dp[1005][1005];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    char a[1005],b[1005];
    while(~scanf(" %s %s",a+1,b+1))
    {
        dp[0][0]=0;
        dp[0][1]=0;
        dp[1][0]=0;
        int len1=strlen(a+1);
        int len2=strlen(b+1);
        for(int i=1;i<=len1;i++)
        {
            for(int j=1;j<=len2;j++)
            {
                if(a[i]==b[j])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
            }
        }
        printf("%d\n",dp[len1][len2]);
    }
    return 0;
}