HDU 1159 Common Subsequence（最长公共子序列）dp

男娘i 2024-02-17 19:25 57阅读 0赞

## 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

4
    2
    0

本题是一个简单的求最长公共子序列个数问题。如不知道解法可访问（http://blog.csdn.net/yysdsyl/article/details/4226630）写的比较详细

AC代码如下;

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=10000;
    int dp[maxn][maxn];
    int main(){
    	char s1[maxn],s2[maxn];
    	while(scanf("%s%s",s1,s2)==2){
    		int len1=strlen(s1);
    		int len2=strlen(s2);
    		for(int i=1;i<=len1;i++)
    		 dp[i][0]=0;
    		for(int i=1;i<=len2;i++)
    		  dp[0][i]=0;
    		  
    		for(int i=1;i<=len1;i++){
    			for(int j=1;j<=len2;j++){
    				if(s1[i-1]==s2[j-1]){
    					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;
    }