OpenJudge/Poj 1458 Common Subsequence

1.链接地址：

http://poj.org/problem?id=1458

http://bailian.openjudge.cn/practice/1458/

2.题目：

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35411		Accepted: 14080

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, x _ij = 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

Southeastern Europe 2003

3.思路：

 

4.代码：

 1 #include "stdio.h"

 2 

 3 //#include "stdlib.h"

 4 

 5 #include "string.h"

 6 

 7 #define N 1000

 8 

 9 char a[N],b[N];

10 

11 int c[N+2][N+2];

12 

13 int dp(int lena,int lenb)

14 

15 {

16 

17     int i,j;

18 

19     for(i=0;i<=lena;i++) {c[i][0]=0;}

20 

21     for(j=1;j<=lenb;j++) {c[0][j]=0;}

22 

23     for(i=1;i<=lena;i++)

24 

25     {

26 

27        for(j=1;j<=lenb;j++)

28 

29        {

30 

31            if(a[i-1] == b[j-1]) {c[i][j]=c[i-1][j-1]+1;}

32 

33            //if(i==1 && j==1){printf("%c %c\n",a[0],b[0]);}

34 

35            else { c[i][j]=(c[i][j-1]>c[i-1][j])?c[i][j-1]:c[i-1][j]; }

36 

37            //printf("%c %c %d\n",a[i-1],b[j-1],c[i][j]);

38 

39        }

40 

41     }

42 

43     return c[lena][lenb];

44 

45 }

46 

47 

48 

49 int main()

50 

51 {

52 

53     int i,j;

54 

55     while(scanf("%s%s",a,b) != EOF)

56 

57     {

58 

59        printf("%d\n",dp(strlen(a),strlen(b)));

60 

61        //for(i=0;i<strlen(a);i++){for(j=0;j<strlen(b);j++){printf("%d ",c[i][j]);}printf("\n");}

62 

63     }

64 

65     //system("pause");

66 

67     return 0;

68 

69 }