HDU1159 && POJ1458:Common Subsequence(LCS)

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

//入门级问题,算法导论上很详细

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

char str1[1005],str2[1005];
int len1,len2;
int dp[1005][1005];

void LCS()
{
    len1=strlen(str1);
    len2=strlen(str2);
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=len1;i++)
    {
        for(int j=1;j<=len2;j++)
        {
            if(str1[i-1]==str2[j-1])
            dp[i][j]=dp[i-1][j-1]+1;
            else
            dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
        }
    }
}

int main()
{
    while(cin>>str1>>str2)
    {
        LCS();
        cout<<dp[strlen(str1)][strlen(str2)]<<endl;
    }
    return 0;
}
时间: 2024-11-16 02:18:31

HDU1159 && POJ1458:Common Subsequence(LCS)的相关文章

POJ - 1458 - Common Subsequence (LCS)

题目传送:Common Subsequence 思路:LCS AC代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <vector> #include <map> #i

HDU 1159:Common Subsequence(最长公共子序列)

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23108    Accepted Submission(s): 10149 Problem Description A subsequence of a given sequence is the given sequence with some e

POJ1458 &amp;&amp; HDOJ1159 Common Subsequence【LCS】

题目链接(POJ) :http://poj.org/problem?id=1458 题目链接(HDOJ):http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 40156   Accepted: 16162 Description A subsequence of a given sequence

【POJ1458】Common Subsequence

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42100   Accepted: 16996 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..

HDU1159 &amp;&amp; POJ1458 Common Subsequence (LCS模版题)

题目大意:求出两个串的公共子序列的长度 LCS的入门题,读懂题了直接模板就可以 #include <iostream> #include <cstring> using namespace std; const int N=1000; int a[N][N]; int LCS(const char *s1, const char *s2) {// s1:0...m, s2:0...n int m = strlen(s1), n = strlen(s2); int i, j; a[0

HDU 1159 Common Subsequence (LCS)

题意:给定两行字符串,求最长公共子序列. 析:dp[i][j] 表示第一串以 i 个结尾和第二个串以 j 个结尾,最长公共子序列,剩下的就简单了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <

Longest Common Subsequence (LCS)

最长公共子序列(LCS)是经典的DP问题,求序列a[1...n], b[1..m]的LCS. 状态是DP[i][j],表示a[1..i],b[1..j]的LCS. DP转移方程是 DP[i][j]= DP[i-1][j-1]+1, a[i] == b[j] max{ DP[i][j-1], DP[i-1][j] }, a[i] != b[i]  ----------------------------------------------------------------------------

动态规划 ---- 最长公共子序列(Longest Common Subsequence, LCS)

分析: 完整代码: // 最长公共子序列 #include <stdio.h> #include <algorithm> using namespace std; const int N = 100; char A[N], B[N]; int dp[N][N]; int main() { freopen("in.txt", "r", stdin); int n; gets(A + 1); // 从下标1开始读入 gets(B + 1); in

【算法导论学习-29】动态规划经典问题02:最长公共子序列问题(Longest common subsequence,LCS)

问题描述:序列X={x1,x2,-,xn},Y={y1,y2,-,yn},当Z={z1,z2-,zn}是X的严格递增下标顺序(可以不连续)的子集,也是Y的严格递增下标顺序(可以不连续)的子集,则Z是X和Y的公共子序列.例如X={A,B,C,B,D,A,B},Y={B,D,C,A,B,A},{B,C,A}.{B,C,B,A}.{B,D,A,B}都是X和Y的公共子序列.其中最长的公共子序列叫做Longest common subsequence,即经典的LCS. 具体点:char[]xArray和c