标签：
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
题意：多组测试实例，每组两个字符串，求出两个字符串最大的一一对应的相等字符的数目（最长公共子序列的长度）。
例如abcfbc 和 abfcab: 它们的最长公共子序列是abfc或abfb，那么它们的最长公共子序列的长度为4
a b c f b c
a 1 1 1 1 1 1
b 1 2 2 2 2 2
f 1 2 2 3 3 3
c 1 2 3 3 3 4
a 1 2 3 3 3 4
b 1 2 3 3 4 4
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=1010;
int dp[N][N];
int main ()
{
int i, j, m, n;
char a[N], b[N];
while (scanf("%s %s", a, b) != EOF)
{
m = strlen(a);
n = strlen(b);
memset(dp, 0, sizeof(dp));
for (i = 1; i <= m; i++)
{
for (j = 1; j <= n; 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[m][n]);
}
return 0;
}
POJ 1458 Common Subsequence
标签：
原文地址：http://www.cnblogs.com/syhandll/p/4732783.html
