POJ 1458 Common Subsequence（最长公共子序列）

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 61531   Accepted: 25742 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. Input The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. Output 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 Source Southeastern Europe 2003

解题思路：

边界条件：  MaxLen(n,0)=0(n=0,1,2,......len1)          MaxLen(0,n)=0(n=0,1,2.......len2)

如果第n个字母相同

         if(s1[i-1]==s2[j-1])         MaxLen(i-1,j-1)+1

否则        Max(MaxLen(i,j-1),MaxLen(i-1,j))   时间复杂度o(mn)

AC代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char sz1[1000];
char sz2[1000];
int maxLen[1000][1000];
int main(){
    while(cin>>sz1>>sz2){
        int len1=strlen(sz1);
        int len2=strlen(sz2);
        int nTmp;
        int i,j;
        for(i=0;i<=len1;i++)
            maxLen[i][0]=0;//边界条件
        for(j=0;j<len2;j++)
            maxLen[0][j]=0;
        for(i=1;i<=len1;i++){
        for(j=1;j<=len2;j++){
            if(sz1[i-1]==sz2[j-1])
                maxLen[i][j]=maxLen[i-1][j-1]+1;
            else
                maxLen[i][j]=max(maxLen[i-1][j],maxLen[i][j-1]);

        }
        }
        cout<<maxLen[len1][len2]<<endl;
    }
    return 0;
}