题目链接：http://poj.org/problem?id=1458
题目大意：最长公共子序列，不知道的话回去看书吧。。。算了， 就是他们共有的长度最大的子序列，可以是不连续的。问最大长度。
解题思路：dp[i][j]:=串A的前i个字符和串B的前j个字符的最长公共子序列长度
if(a[i] == b[j])
dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
else
dp[i][j] = max(dp[i][j], max(dp[i - 1][j], dp[i][j - 1]));
代码：
1 const int maxn = 1e3 + 5; 2 char a[maxn], b[maxn]; 3 int dp[maxn][maxn]; 4 5 int solve(){ 6 memset(dp, 0, sizeof(dp)); 7 int asz = strlen(a + 1), bsz = strlen(b + 1); 8 for(int i = 1; i <= asz; i++){ 9 for(int j = 1; j <= bsz; j++){ 10 if(a[i] == b[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1); 11 else dp[i][j] = max(dp[i][j], max(dp[i - 1][j], dp[i][j - 1])); 12 } 13 } 14 printf("%d ", dp[asz][bsz]); 15 return 0; 16 } 17 int main(){ 18 while(scanf("%s %s", a + 1, b + 1) != EOF){ 19 solve(); 20 } 21 }
题目：
Common Subsequence
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 54425 | Accepted: 22632 |
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