HDOJ 1159 Common Subsequence【DP】

HDOJ1159最长公共子序列解析

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本文详细介绍了如何通过动态规划解决HDOJ1159最长公共子序列问题，并提供了完整的AC代码实现。文章首先定义了最长公共子序列的概念，接着阐述了解题的基本思路：当两字符串末尾字符相等，则当前最长公共子序列长度为前一状态长度加一；若不相等则取两者最大值。

HDOJ 1159 Common Subsequence【DP】

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44280 Accepted Submission(s): 20431

Problem 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, 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.

Sample Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4
2
0

题意

求解两个字符串的最长公共子序列

思路
如果两个字符串的最后一个字符相等，那么由这最后一个字符组成的最长公共子序列就是前面的最长公共子序列长度+ 1 然后往前推就可以了
DP[i][j] = DP[i - 1][j - 1] + 1
如果不相等
DP[i][j] = max(DP[i - 1][j], DP[i][j - 1])

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <cstdlib>
#include <ctype.h>
#include <numeric>
#include <sstream>
using namespace std;

typedef long long LL;
const double PI = 3.14159265358979323846264338327;
const double E = 2.718281828459;
const double eps = 1e-6;
const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e3 + 5;
const int MOD = 1e9 + 7;
int dp[maxn][maxn];

int main()
{
    string a, b;
    while (cin >> a >> b)
    {
        int len_a = a.size(), len_b = b.size();
        memset(dp, 0, sizeof(dp));
        LL ans = 0;
        for (int i = 0; i < len_a; i++)
        {
            if (b[0] == a[i])
            {
                dp[i][0] = 1;
                ans = 1;
            }
            else if (i)
                dp[i][0] = dp[i - 1][0];
        }
        for (int i = 0; i < len_b; i++)
        {
            if (a[0] == b[i])
            {
                dp[0][i] = 1;
                ans = 1;
            }
            else if(i)
                dp[0][i] = dp[0][i - 1];
        }
        for (int i = 1; i < len_a; i++)
        {
            for (int j = 1; j < len_b; 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]);
            }
        }
        cout << dp[len_a - 1][len_b - 1] << endl;
    }
}