poj 1458 Common Subsequence

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 47381 Accepted: 19464

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

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【分析】
突然发现刷水题成为了我的乐趣。。
这道题就是求最长公共子序列
dp[i][j]表示a串的前i个字符和b串的前j个字符所能组成的最长公共子序列长度
然后推出状态转移方程（具体见代码）

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【代码】

//poj 1458 Common Subsequence#include<iostream>#include<cstdio>#include<cstring>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;string a,b;int dp[1001][1001];int main(){    int i,j,k;    while(cin>>a>>b)    {        memset(dp,0,sizeof dp);        int la=a.size()-1;        int lb=b.size()-1;        fo(i,0,la)          fo(j,0,lb)          {            if(a[i]==b[j])              dp[i+1][j+1]=dp[i][j]+1;            else              dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);          }        printf("%d\n",dp[la+1][lb+1]);    }    return 0;}

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【附上另一份脑残代码】
【唯一不同就是下标的不同，这里的dp[i][j]表示i之前与j之前的串】

//poj 1458 Common Subsequence#include<iostream>#include<cstdio>#include<cstring>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;string a,b;int dp[1001][1001];int main(){    int i,j,k;    while(cin>>a>>b)    {        memset(dp,0,sizeof dp);        int la=a.size()-1;        int lb=b.size()-1;        fo(i,0,la)          fo(j,0,lb)          {            if(a[i]==b[j])              if(i-1>=0 && j-1>=0)                dp[i][j]=dp[i-1][j-1]+1;              else dp[i][j]=1;            else            {                if(i-1>=0)                  dp[i][j]=max(dp[i-1][j],dp[i][j]);                if(j-1>=0)                  dp[i][j]=max(dp[i][j-1],dp[i][j]);            }          }        printf("%d\n",dp[la][lb]);    }    return 0;}