先上题目：
Common Subsequence
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 19679 Accepted
Submission(s): 8389
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
题意就是求两个字符串的最长公共子串的长度。这题可以算是LCS的例题=
=，所以直接写一个求两个字符串的LCS就可以了，但是这里没有告诉我们字符串的长度，所以字符串数组开了1000+，所以用来记录LCS的数组开不了这么大，只好用滚动数组，在原本的LCS上面稍微改一下就可以了。
上代码：
1 #include <cstdio> 2 #include <cstring> 3 #define MAX 1100 4 using namespace std; 5 6 char a[MAX],b[MAX]; 7 int c[11][MAX]; 8 9 int main() 10 { 11 int lena,lenb,maxn,v; 12 while(scanf("%s %s",a+1,b+1)!=EOF){ 13 lena=strlen(a+1); 14 lenb=strlen(b+1); 15 memset(c,0,sizeof(c)); 16 maxn=0; 17 v=1; 18 for(int i=1;i<=lena;i++){ 19 v=i%10; //这一句需要放在外面的循环里 20 for(int j=1;j<=lenb;j++){ 21 if(a[i]==b[j]) c[v][j]=c[(v-1+10)%10][j-1]+1; 22 else if(c[v][j-1]>=c[(v-1+10)%10][j]){ 23 c[v][j]=c[v][j-1]; 24 }else{ 25 c[v][j]=c[(v-1+10)%10][j]; 26 } 27 maxn = c[v][j] > maxn ? c[v][j] : maxn; 28 } 29 } 30 printf("%d\n",maxn); 31 32 } 33 return 0; 34 } 35 36 37 /** 38 abcfbc abfcab 39 programming contest 40 abcd mnp 41 42 4 43 2 44 0 45 */