【HDU 1159】 Common Subsequence（dp —lcs模板题）

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40637    Accepted Submission(s): 18752

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 abfcabprogramming contest abcd mnp

 

Sample Output

420

 

Source

Southeastern Europe 2003

 

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题意：求最大的公共子序列

题解：lcs模板题

代码：

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;const int MAX=1e3+10;int dp[MAX][MAX];char str1[MAX],str2[MAX];int main(){while(~scanf("%s %s",str1+1,str2+1)){str1[0]=str2[0]='0';memset(dp,0,sizeof(dp));int len1=strlen(str1)-1;int len2=strlen(str2)-1;for(int i=1;i<=len1;i++){for(int j=1;j<=len2;j++){if(str1[i] == str2 [j])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);}}cout<<dp[len1][len2]<<endl;}}

附：lcs算法

还可以回溯求最长公共子序列元素：

代码：

/*LCSBDCABAABCBDABdp[1][2] = 1dp[1][1] = 0dp[2][1] = 0//子串：连续//子序列：可以不连续 // LCSdp[i][j]//第一个字符串在第i个字符前且第二个串在第j个字符前可构成的最长子序列的长度 dp[i][j] = 0  i=0 || j=0dp[i-1][j-1]+1     str1[i]==str2[j]max(dp[i-1][j],dp[i][j-1]) str1[i]!=str2[j]*/#include<cstdio>#include<cstring>#include<stack>#include<algorithm>using namespace std;int main(){char str1[20];char str2[20];scanf ("%s %s",str1+1,str2+1);str1[0] = str2[0] = '0';int l1 = strlen(str1)-1;int l2 = strlen(str2)-1;int dp[20][20] = {0};//0  i=0 || j=0for (int i = 1 ; i <= l1 ; i++){for (int j = 1 ; j <= l2 ; j++){if (str1[i] == str2[j])//dp[i-1][j-1]+1     str1[i]==str2[j]dp[i][j] = dp[i-1][j-1] + 1;elsedp[i][j] = max(dp[i-1][j],dp[i][j-1]);//max(dp[i-1][j],dp[i][j-1]) str1[i]!=str2[j]}}//回溯求LCS int pos1 = l1;int pos2 = l2;stack<char> S;while (pos1 > 0 && pos2 > 0){if (str1[pos1] == str2[pos2]){S.push(str1[pos1]);pos1--;pos2--;}else if (dp[pos1-1][pos2] > dp[pos1][pos2-1])pos1--;elsepos2--;}while (!S.empty()){printf ("%c%c",S.top(),(S.size() == 1) ? '\n' : ' ');S.pop();}printf ("%d\n",dp[l1][l2]);return 0;}