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ZOJ 1733 Common Subsequence(LCS)

Common Subsequence

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

分析:最长公共子序列

代码如下:

 

 1 # include<stdio.h>
 2 # include<string.h>
 3 # define MAX 1005
 4 char s1[MAX],s2[MAX];
 5 int dp[MAX][MAX];
 6 int len1,len2;
 7 int max(int a,int b,int c){
 8     int temp;
 9     temp = a>b ? a : b;
10     return temp>c ? temp : c;
11 }
12 int main(){
13     int i,j;
14     while(scanf("%s%s",s1,s2)!=EOF){
15         len1 = strlen(s1);
16         len2 = strlen(s2);
17         memset(dp,0,sizeof(dp));
18         for(i=1;i<=len1;i++){
19             for(j=1;j<=len2;j++){
20                 if(s1[i-1] == s2[j-1])
21                     dp[i][j] = dp[i-1][j-1] + 1;
22                 dp[i][j] = max(dp[i][j],dp[i-1][j],dp[i][j-1]);
23             }
24         }
25         printf("%d\n",dp[len1][len2]);
26     }
27     return 0;
28 }

 

 

 

posted @ 2013-08-15 02:40  贾树丙  阅读(309)  评论(0编辑  收藏  举报