POJ-1458 Common Subsequence 

 1 #include <iostream>
 2 #include <string>
 3 
 4 //求最长公共子序列
 5 using namespace std;
 6 
 7 int const MAX_NUM = 1010;
 8 int dp[MAX_NUM][MAX_NUM];
 9 
10 int max(int a,int b){
11     return a>b?a:b;
12 }
13 
14 int main(){
15     
16     string s1,s2;
17     while(cin>>s1>>s2){
18         int len1 = s1.length();
19         int len2 = s2.length();
20 
21         for(int i = 1; i <= len1; i++){
22             for(int j = 1; j <= len2; j++){
23                 dp[i][j] = 0;
24             }
25         }
26 
27         for(int i = 1; i <= len1; i++){
28             for(int j = 1; j <= len2; j++){
29                 if(s1[i-1] == s2[j-1]){
30                     dp[i][j] = dp[i-1][j-1]+1;
31                 }else{
32                     dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
33                 }
34             }
35         }
36 
37         cout<<dp[len1][len2]<<endl;
38     }
39 
40     return 0;
41 }

 分析：采用动态规划解决

posted @ 2016-06-12 11:29  hu983  阅读(170)  评论(0)  编辑  收藏  举报