C - Common Subsequence

C - Common Subsequence

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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. 

Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0


//求两个字符串的最长公共子序列长度

//虽然是 dp 序列水题，但是我第一次做不会，没想到转移方程

代码里写的很清楚了，31ms

dp[i][j]表示0到i-1跟0到j-1的最长公共子序列长度

#include <stdio.h>
#include <string.h>

char a[1005];
char b[1005];
int dp[1005][1005];

int max(int x,int y)
{return x>y?x:y;}

int main()
{
    int i,j;
    while(scanf("%s%s",a,b)!=EOF)
    {
        int la=strlen(a),lb=strlen(b);
        for (i=0;i<=lb;i++)
            dp[0][i]=0;
        for (i=0;i<=la;i++)
            dp[i][0]=0;
        for (i=1;i<=la;i++)
        {
            for (j=1;j<=lb;j++)
            {
                if (a[i-1]==b[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[la][lb]);
    }
    return 0;
}