HDU 1159 Common Subsequence(POJ 1458)

Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34819 Accepted Submission(s): 15901

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.
Sample Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4
2
0

Source Southeastern Europe 2003
解析：最长公共子序列。
``` #include #include #include using namespace std;

char s1[500], s2[500];
int dp[500][500];

int lcs()
{
int m = strlen(s1), n = strlen(s2);
for(int i = 0; i <= m; ++i)
dp[i][0] = 0;
for(int i = 0; i <= n; ++i)
dp[0][i] = 0;
for(int i = 1; i <= m; ++i){
for(int j = 1; j <= n; ++j){
if(s1[i-1] == s2[j-1])
dp[i][j] = dp[i-1][j-1]+1;
else
dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
}
}
return dp[m][n];
}

int main()
{
while(~scanf("%s%s", s1, s2)){
int res = lcs();
printf("%d ", res);
}
return 0;
}

<br>
因为进行状态转移的时候，只与前一个状态有关，所以可以用[滚动数组](http://blog.csdn.net/insistgogo/article/details/8581215)进行优化来减少所需存储空间。

include

include

include

using namespace std;

char s1[500], s2[500];
int dp[2][500];

int lcs()
{
int m = strlen(s1), n = strlen(s2);
memset(dp, 0, sizeof dp);
for(int i = 1; i <= m; ++i){
for(int j = 1; j <= n; ++j){
if(s1[i-1] == s2[j-1])
dp[i%2][j] = dp[(i-1)%2][j-1]+1;
else
dp[i%2][j] = max(dp[i%2][j-1], dp[(i-1)%2][j]);
}
}
return dp[m%2][n];
}

int main()
{
while(~scanf("%s%s", s1, s2)){
int res = lcs();
printf("%d ", res);
}
return 0;
}