Common Subsequence 动态规划I

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roblem 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 subsequenceof 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.

Sample Input
abcfbc abfcab
programming contest
abcd mnp

Sample Output
4
2
0

Source
Southeastern Europe 2003

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题意：对于母串X=<x1,x2,......,xm>, Y=<y1,y2,......,yn>，求最长公共子串。

思路：用二维数组dp[i][j]记录串x1x2......xi与y1y2......yj的能得到最长公共子串，则可得到状态转移方程
当a[i] == a[j] 时 dp[i][j] = dp[i-1][j-1] + 1;
否则 dp[i][j] = max(dp[i][j-1],dp[i-1][j]);

#include<iostream>#include<cstring>#include<string.h>#include<algorithm>//max()的函数const int N = 1000 + 24;char a[N], b[N];int dp[N][N];using namespace std;int main(){#ifdef CDZSC_OFFLINE;freopen("in.txt", "r", "stbin");#endif;while (scanf("%s %s", a + 1, b + 1) != EOF)//注意此处输入前是%s,当输入字符串数组时前为%s,a+1,b+1 表示输入的数组从第二项（1）开始标志，是为了下面从1开始循环方便{int i, j;memset(dp, 0, sizeof(dp));int len1 = strlen(a + 1);//计算字符串的长度int len2 = strlen(b + 1);for (i = 1; i <= len1; i++)//此处的思想是将数组a中的每个元素逐次和数组b比较for (j = 1; j <= len2; j++){if (a[i] == b[j])//如果a中的第i个元素等于b中的第j个元素dp[i][j] = dp[i - 1][j - 1] + 1;//dp[i][j]左上方那个数+1，然后动态转移elsedp[i][j] = max(dp[i][j-1], dp[i - 1][j]);//选出dp[i][j]左侧或者上面最大的那个值动态转移}printf("%d\n", dp[len1][len2]);//数组二维数组最后一个元素的值}}

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