【北大夏令营笔记-动态规划】poj1458-Common Subsequence

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 37543 Accepted: 15016

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

//题意：最长公共子序列

/*

输入两个串s1，s2.设MaxLen[i][j]表示：

s1的左边i个字符形成的子串，与s2左边的j个字符形成的子串的最长公共子

序列的长度（i，j从0开始计算）。

MaxLen[i][j]就是本题的“状态”。

*/
AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=1100;
char s1[MAXN];
char s2[MAXN];
int MaxLen[MAXN][MAXN];
int main()
{
int i,j,n,m;
while(scanf("%s %s",s1,s2)!=EOF)
{
n=strlen(s1);
m=strlen(s2);
for(i=0;i<=n;i++)
MaxLen[i][0]=0;
for(i=0;i<=m;i++)
MaxLen[0][i]=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
if(s1[i-1]==s2[j-1])
MaxLen[i][j]=MaxLen[i-1][j-1]+1;
else
MaxLen[i][j]=max(MaxLen[i][j-1],MaxLen[i-1][j]);
}
printf("%d\n",MaxLen
[m]);
}
return 0;
}