POJ1458 Common Subsequence

题目链接：http://poj.org/problem?id=1458

分析：最大公共子序列模板

 1 #include<iostream>
 2 #include<sstream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<string>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<functional>
 9 #include<iomanip>
10 #include<numeric>
11 #include<cmath>
12 #include<queue>
13 #include<vector>
14 #include<set>
15 #include<cctype>
16 const double PI = acos(-1.0);
17 const int INF = 0x3f3f3f3f;
18 const int NINF = -INF - 1;
19 const int maxn = 1e3 + 5;
20 typedef long long ll;
21 #define MOD 1000000007
22 using namespace std;
23 char a[maxn], b[maxn];
24 int dp[maxn][maxn];
25 int main()
26 {
27     while (scanf("%s", &a) != EOF)
28     {
29         scanf("%s", &b);
30         memset(dp, 0, sizeof(dp));
31         int len1 = strlen(a);
32         int len2 = strlen(b);
33         for (int i = 1; i <= len1; ++i)
34         {
35             for (int j = 1; j <= len2; ++j)
36             {
37                 if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
38                 else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
39             }
40         }
41         printf("%d\n", dp[len1][len2]);
42     }
43     return 0;
44 }

 

posted @ 2019-09-21 15:45  Veasky  阅读(162)  评论(0)  编辑  收藏  举报