Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40883 Accepted Submission(s): 18866
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 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.
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
题解：
给出两个字符串，找出一个最长的公共子序列（子序列不一定连续），求子序列的长度
这个是我自己的例子
s1:coprmmt s2:progrpmming
当 s1[i] == s2[j]时，则在之前的子序列最长的那个加上1（dp[i][j] = dp[i-1][j-1]+1）
给出的转移方程为
if(s1[i] == s2[j])
dp[i][j] = dp[i-1][j-1]+1;
else
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
同时还要注意边界情况，所以在第0行，0列初始值都为0，行和列都增加一行或一列
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 int dp[1100][1100]; 7 int main() 8 { 9 char s1[1500],s2[1500]; 10 while(scanf("%s%s",s1,s2) != EOF) { 11 int len1 = strlen(s1); 12 int len2 = strlen(s2); 13 for(int i = 0; i < len1; i++) 14 dp[i][0] = 0; 15 for(int i = 0; i < len2; i++) 16 dp[0][i] = 0; 17 for(int i = 0; i < len1; i++) { 18 for(int j = 0; j < len2; j++) { 19 if(s1[i] == s2[j]) dp[i+1][j+1] = dp[i][j]+1; 20 else dp[i+1][j+1] = dp[i+1][j] > dp[i][j+1]? dp[i+1][j]:dp[i][j+1]; 21 } 22 } 23 printf("%d ",dp[len1][len2]); 24 } 25 return 0; 26 }