UVA 10405 Longest Common Subsequence

Posted 2020-09-22

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=16&page=show_problem&problem=1346

Dynamic programming

需要注意的是input里可能含有空行，一行空行也是一个string，所以如果用scanf("%s", string)是不能把空行存进string变量里的。需要用gets或者getline来处理input。

可惜的是我目前对于gets，getline的用法还没有很熟悉，之后会关于这方面的区别做解释。又或者有大神可以帮忙讲解一下，感恩

代码如下：

 1 //============================================================================
 2 // Name        : test.cpp
 3 // Author      : 
 4 // Version     :
 5 // Copyright   : Your copyright notice
 6 // Description : Hello World in C++, Ansi-style
 7 //============================================================================
 8 
 9 #include <iostream>
10 #include <math.h>
11 #include <stdio.h>
12 #include <cstdio>
13 #include <algorithm>
14 #include <string.h>
15 #include <cstring>
16 #include <queue>
17 #include <vector>
18 #include <functional>
19 #include <cmath>
20 #define SCF(a) scanf("%d", &a)
21 #define IN(a) cin>>a
22 #define FOR(i, a, b) for(int i=a;i<b;i++)
23 typedef long long Int;
24 using namespace std;
25 
26 int main()
27 {
28     char str1[1005];
29     char str2[1005];
30     int len1, len2;
31     while (cin.getline(str1, 1005))
32     {
33         cin.getline(str2, 1005);
34         for (len1 = 0; str1[len1] != ‘\0‘; len1++);
35         for (len2 = 0; str2[len2] != ‘\0‘; len2++);
36         int** match;
37         match = new int*[len1+1];
38         FOR(i, 0, len1 + 1)
39             match[i] = new int[len2 + 1];
40         FOR(i, 0, len1 + 1)
41             match[i][0] = 0;
42         FOR(i, 0, len2 + 1)
43             match[0][i] = 0;
44 
45         FOR(i, 1, len1 + 1)
46         {
47             FOR(j, 1, len2 + 1)
48             {
49                 if (str1[i - 1] == str2[j - 1])
50                     match[i][j] = match[i - 1][j - 1] + 1;
51                 else
52                 {
53                     match[i][j] = max(match[i - 1][j], match[i][j - 1]);
54                 }
55             }
56         }
57         
58         printf("%d\n", match[len1][len2]);
59         FOR(i, 0, len1 + 1)
60             delete[] match[i];
61         delete[]  match;
62     }
63     return 0;
64 }