2402: Common Subsequence
时间限制: 1 Sec 内存限制: 32 MB提交: 63 解决: 33
题目描述
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.
输入
abcfbc abfcab
programming contest
abcd mnp
输出
4
2
0
样例输入
abcfbc abfcab
programming contest
abcd mnp
样例输出
4
2
0
迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char s1[1000],s2[1000];
int dp[1000][1000];
int len1,len2;
void LCS()
{
int i,j;
memset(dp,0,sizeof(dp));
for(i = 1; i<=len1; i++)
{
for(j = 1; j<=len2; j++)
{
if(s1[i-1] == s2[j-1])
dp[i][j] = dp[i-1][j-1]+1;
else
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
}
int main()
{
while(~scanf("%s%s",s1,s2))
{
len1 = strlen(s1);
len2 = strlen(s2);
LCS();
printf("%d
",dp[len1][len2]);
}
return 0;
}