HDUOJ 1159

原题：

• Problem 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.
  The program…….

• Sample Input

  abcfbc abfcab
  programming contest
  abcd mnp

• Sample Output

  4
  2
  0

解题思路：

• 可以把c[ i ][ j ]理解成：取X的前i个字符和取Y的前j个字符，能得到的LCS值
• 分为一下两类：
  
  • 如果X的第i个字符X[ i ] 等于 Y的第j个字符Y[ j ]
  • 如果X的第i个字符X[ i ] 不等于 Y的第j个字符Y[ j ]

引进一个二维数组c[ ][ ]，用c[ i ][ j ]记录X[ i ]与Y[ j ] 的LCS 的长度。
我们是自底向上从小到大进行递推计算，所以就能在在计算c[ i ][ j ]之前，把c[ i-1 ][ j-1 ]，c[ i-1 ][ j ]与c[ i ][ j-1 ]计算出来，此时我们根据X[ i ] = Y[ j ]还是X[ i ] != Y[ j ]，就可以计算出c[ i ][ j ]。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int dp[1000][1000];char a[1000], b[1000];int main(){    int a_length, b_length, maxLength, i, j;    while (scanf("%s%s",a + 1,b + 1)!=EOF)//0位不存    {        a[0] = b[0] = '*';//无用--只是方便下面数组循环无需减一加一操作        a_length = strlen(a);        b_length = strlen(b);        maxLength = max(a_length, b_length);        for (i = 0; i <= maxLength; i++)            dp[i][0] = dp[0][i] = 0;//初始化        for (i = 1; i < a_length; i++)//动态规划            for (j = 1; j < b_length; j++)                if (a[i] == b[j])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[i - 1][j - 1]);    }}