/*
dp[i][k] := 前i+1个字符组成的字符串，划分为k份，每份中包含的单词个数加起来总数的最大值。
初始化：
dp[][] = -0x3f3f3f3f
// 注意：对于dp[i][k]，若(i+1)<k，则dp[i][k]的值不存在，设为-0x3f3f3f3f。例如:dp[0][2] = -0x3f3f3f3f
dp[i][1] = val[0][i]
//注：val[i][j] := 截取原字符串【从i到j的字符（包含i和j位置的字符）】组成字符串，返回其包含的单词个数
状态方程：
dp[i][k] = max(dp[i][k], dp[j][k-1]) + Num(j+1, i))
(0<=j<i)
答案：
dp[len-1][k]
此题难点：求val[i][j]
val[i][j] = val[i][j-1] + num(j)
num(j) := 遍历所有想要匹配的字符串strArr[m]，从原字符串strS位置j出发向前找字符，找到一个首字符位置fst,
使得strS[fst,j]的长度刚好等于strArr[m],若满足以下条件：
1）i<=fst<=j
对应代码：(j - i + 1) >= strArr[m].size()
2）此字符串首字符的位置fst在求val[i][j-1]和求num(j)时，没有用过
对应代码：!used[j - strArr[m].size() + 1]
3）strS[i,j] == strArr[m]
对应代码：strSource.substr(j - strArr[m].size() + 1, strArr[m].size()) == strArr[m]
则：+1
最后找到几个满足这三个条件的首位置fst，num(j)的值就是几。
*/
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <cstddef>
5 #include <iterator>
6 #include <algorithm>
7 #include <string>
8 #include <locale>
9 #include <cmath>
10 #include <vector>
11 #include <cstring>
12 using namespace std;
13 const int INF = -0x3f3f3f3f;
14 const int MaxN = 210;
15 const int MaxK = 50;
16
17 int dp[MaxN][MaxK];
18 int val[MaxN][MaxN];
19 int len, K, S;
20 string strSource;
21 string strArr[10];
22
23
24 void iniVal()
25 {
26 bool used[MaxN];
27 for (int i = 0; i < len; ++i)
28 {
29 memset(used, 0, sizeof(used));
30 for (int j = i; j < len; ++j)
31 {
32 if (i != j) val[i][j] = val[i][j - 1];
33 for (int m = 0; m < S; ++m)
34 {
35 if ((j - i + 1 >= strArr[m].size()) && !used[j - strArr[m].size() + 1]
36 && (strSource.substr(j - strArr[m].size() + 1, strArr[m].size()) == strArr[m]))
37 {
38 ++val[i][j];
39 used[j - strArr[m].size() + 1] = true;
40 }
41 }
42 }
43 }
44 }
45
46 void Solve()
47 {
48 for (int i = 0; i < len; ++i)
49 {
50 dp[i][1] = val[0][i];
51 }
52 for (int i = 0; i < len; ++i)
53 {
54 for (int j = 0; j < i; ++j)
55 {
56 for (int k = 2; k <= K; ++k)
57 {
58 dp[i][k] = max(dp[i][k], dp[j][k - 1] + val[j+1][i]);
59 int bbb = 1;
60 }
61 }
62 }
63 cout << dp[len-1][K] << endl;
64
65
66
67 /*cout << "\n\n\n\n" << "val:----------------------------" << endl;
68 for (int i = 0; i < len; ++i)
69 {
70 for (int j = 0; j < len; ++j)
71 {
72 cout << val[i][j] << " ";
73 }
74 cout << endl;
75 }
76
77 cout << "\n\n\n\n" << "dp:----------------------------" << endl;
78 for (int i = 0; i < len; ++i)
79 {
80 for (int j = 1; j <= K; ++j)
81 {
82 cout << dp[i][j] << " ";
83 }
84 cout << endl;
85 }*/
86 }
87
88 int main()
89 {
90 #ifdef HOME
91 freopen("in", "r", stdin);
92 //freopen("out", "w", stdout);
93 #endif
94
95 int cas;
96 cin >> cas;
97 while (cas--)
98 {
99 strSource = "";
100 memset(val, 0, sizeof(val));
101 for (int i = 0; i < MaxN; ++i)
102 {
103 for (int j = 0; j < MaxK; ++j)
104 {
105 dp[i][j] = INF;
106 }
107 }
108 int p;
109 cin >> p >> K;
110 string strTmp;
111 for (int i = 0; i < p; ++i)
112 {
113 cin >> strTmp;
114 strSource += strTmp;
115 }
116 len = strSource.size();
117 cin >> S;
118 for (int i = 0; i < S; ++i)
119 {
120 cin >> strArr[i];
121 }
122 iniVal();
123 Solve();
124 }
125
126 #ifdef HOME
127 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
128 _CrtDumpMemoryLeaks();
129 system("pause");
130 #endif
131 return 0;
132 }
