hdu1159

maksyuki 发表于 oj 分类，标签: DP-线性DP

15 9月 2015

Common Subsequence

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

Source

Southeastern Europe 2003

Recommend

Ignatius

题目类型：线性DP

算法分析：最长公共子序列的模板题，直接求解即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
const int MOD = 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 2000 + 66;

int dp[maxn][maxn];
char sa[maxn], sb[maxn];

int main()
{
//	ifstream cin ("aaa.txt");
//	freopen ("aaa.txt", "r", stdin);

    while (cin>> (sa + 1) >> (sb + 1))
	{
        int lena = strlen (sa + 1), lenb = strlen(sb + 1);
        for (int i = 0; i <= lena; i++)
            dp[i][0] = 0;

		for (int i = 0; i <= lenb; i++)
			dp[0][i] = 0;

		for (int i = 1; i <= lena; i++)
		{
			for(int j = 1; j <= lenb; j++)
			{
                if (sa[i] == sb[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[lena][lenb] << endl;
	}

return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

const int MOD = 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 2000 + 66;

int dp[maxn][maxn];

char sa[maxn], sb[maxn];

int main()

{

// ifstream cin ("aaa.txt");

// freopen ("aaa.txt", "r", stdin);

while (cin>> (sa + 1) >> (sb + 1))

{

int lena = strlen (sa + 1), lenb = strlen(sb + 1);

for (int i = 0; i <= lena; i++)

dp[i][0] = 0;

for (int i = 0; i <= lenb; i++)

dp[0][i] = 0;

for (int i = 1; i <= lena; i++)

{

for(int j = 1; j <= lenb; j++)

{

if (sa[i] == sb[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[lena][lenb] << endl;

}

return 0;

}

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Common Subsequence

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