POJ_1458_Common Subsequence

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
1458

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, x ij = 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

这是一个最长公共子序列的问题 设d(i,j)为Ai和Bj的最长公共子序列长度，当a[i]==b[j]时，d(i,j)=d(i-1,j-1)+1, else d(i,j)=max{d(i-1,j),d(i,j-1)}

#include <iostream>

#include <string.h>

#include <stdio.h>

using namespace std;

char a[1200],b[1200];

int dp[1300][1300];

int main()

{

while(scanf("%s%s",a+1,b+1)!=EOF)

{

memset(dp,0,sizeof(dp));

int la=strlen(a+1);

int lb=strlen(b+1);

for(int i=1;i<=la;i++)

{

for(int j=1;j<=lb;j++)

{

if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;

else

{

dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1];

}

}

}

printf("%d\n",dp[la][lb]);

}

return 0;

}