POJ动态规划练习（计划6-8题）

Problems

POJ2192 - Zipper

Solutions

POJ2192 - Zipper

题目大意：给定字符串s1, s2和s，问s1和s2是否满足：均为s的子序列（不必连续）且恰能构成s？

用布尔变量 dp[i][j] 表示 s1 中前 i 个字符和 s2 中前 j 个字符能否构成 s 的前 i + j 个字符构成的子串 s‘，显然 s‘ 的最后一个字符只能是 s1[i] 和 s2[j] 中的一个，由此可以写出状态转移方程。边界条件为 dp[0][0] = 1。

 1 //  Problem: poj2192 - Zipper
 2 //  Category: Dynamic programming
 3 //  Author: Niwatori
 4 //  Date: 2016/07/22
 5
 6 #include <iostream>
 7 #include <cstring>
 8 #include <string>
 9 using namespace std;
10
11 int main()
12 {
13     int t; cin >> t;
14     for (int i = 1; i <= t; ++i)
15     {
16         string s1, s2, s;
17         cin >> s1 >> s2 >> s;
18         int len1 = s1.length(), len2 = s2.length();
19         s1 = ‘ ‘ + s1; s2 = ‘ ‘ + s2; s = ‘*‘ + s;  // For convenience
20
21         bool dp[205][205];
22         memset(dp, 0, sizeof(dp));
23         dp[0][0] = 1;
24         for (int i = 0; i <= len1; ++i)
25             for (int j = 0; j <= len2; ++j)
26             {
27                 if (s1[i] == s[i + j])
28                     dp[i][j] = dp[i][j] || dp[i - 1][j];
29                 if (s2[j] == s[i + j])
30                     dp[i][j] = dp[i][j] || dp[i][j - 1];
31             }
32         cout << "Data set " << i << ": " << (dp[len1][len2] ? "yes" : "no") << endl;
33     }
34     return 0;
35 }