周赛F题 POJ 1458（最长公共子序列）

F - F
Time Limit:1000MS Memory Limit:10000KB 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, 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

题解：给你两个字符串，找到他们最长的公共子序列。这个题做过很多次，虽然加了点东西，还是没有写对，还是对LCS理解不够。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a[1100],b[1100];
int dp[1100][1100];
int main()
{
while(scanf("%s %s",a,b)!=EOF)
{
int x=strlen(a);
int y=strlen(b);
for(int i=1;i<=x;i++)
{
for(int j=1;j<=y;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]);
}
}
cout<<dp[x][y]<<endl;
}
return 0;
}