题意：
就是现在给出m个串，每个串都有一个权值，现在你要找到一个长度不超过n的字符串，
其中之前的m个串每出现一次就算一次那个字符串的权值，
求能找到的最大权值的字符串，如果存在多个解，输出最短的字典序最小的串。
当最大全权值为0时输出空串。
输入最多100个子串，权值为不超过100的正整数。
每个子串长度至少为1，不超过10, n <= 50
如果不考虑方案输出，这题就变得相当简单了。
dp【i】【j】表示走到长度为 i 的时候 ，到AC自动机 j 这个节点所获得的最大权值和。
我一开始的做法是在dp的过程中获得那个最优方案。
最后死活过不去，就换了一种超级暴力的写法。
将所有情况保存下来，去一个个找字典序最小的方案。
1 #include <set>
2 #include <map>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <cstdio>
7 #include <string>
8 #include <vector>
9 #include <time.h>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13
14 #define pi acos(-1.0)
15 #define eps 1e-9
16 #define fi first
17 #define se second
18 #define rtl rt<<1
19 #define rtr rt<<1|1
20 #define bug printf("******\n")
21 #define mem(a, b) memset(a,b,sizeof(a))
22 #define name2str(x) #x
23 #define fuck(x) cout<<#x" = "<<x<<endl
24 #define sf(n) scanf("%d", &n)
25 #define sff(a, b) scanf("%d %d", &a, &b)
26 #define sfff(a, b, c) scanf("%d %d %d", &a, &b, &c)
27 #define sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define pf printf
29 #define FIN freopen("../date.txt","r",stdin)
30 #define gcd(a, b) __gcd(a,b)
31 #define lowbit(x) x&-x
32 #define IO iOS::sync_with_stdio(false)
33
34
35 using namespace std;
36 typedef long long LL;
37 typedef unsigned long long ULL;
38 const int maxn = 1e6 + 7;
39 const int maxm = 8e6 + 10;
40 const int INF = 0x3f3f3f3f;
41 const int mod = 20090717;
42
43
44 char str[105][12];
45 int n, m, cost[105];
46
47
48 struct Aho_Corasick {
49 int next[1010][26], fail[1010], End[1010];
50 int root, cnt;
51
52 int newnode() {
53 for (int i = 0; i < 26; i++) next[cnt][i] = -1;
54 End[cnt++] = 0;
55 return cnt - 1;
56 }
57
58 void init() {
59 cnt = 0;
60 root = newnode();
61 }
62
63 void insert(char buf[], int id) {
64 int len = strlen(buf);
65 int now = root;
66 for (int i = 0; i < len; i++) {
67 if (next[now][buf[i] - 'a'] == -1) next[now][buf[i] - 'a'] = newnode();
68 now = next[now][buf[i] - 'a'];
69 }
70 End[now] += id;
71 }
72
73 void build() {
74 queue<int> Q;
75 fail[root] = root;
76 for (int i = 0; i < 26; i++)
77 if (next[root][i] == -1) next[root][i] = root;
78 else {
79 fail[next[root][i]] = root;
80 Q.push(next[root][i]);
81 }
82 while (!Q.empty()) {
83 int now = Q.front();
84 Q.pop();
85 End[now] += End[fail[now]];
86 for (int i = 0; i < 26; i++)
87 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
88 else {
89 fail[next[now][i]] = next[fail[now]][i];
90 Q.push(next[now][i]);
91 }
92 }
93 }
94
95 void debug() {
96 for (int i = 0; i < cnt; i++) {
97 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
98 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
99 printf("]\n");
100 }
101 }
102 } ac;
103
104 int dp[55][1010];
105 string path[55][1010];
106
107 int main() {
108 // FIN;
109 int T;
110 sf(T);
111 while (T--) {
112 sff(n, m);
113 for (int i = 0; i < m; ++i) scanf("%s", str[i]);
114 for (int i = 0; i < m; i++) sf(cost[i]);
115 ac.init();
116 for (int i = 0; i < m; ++i) ac.insert(str[i], cost[i]);
117 ac.build();
118 mem(dp, -1);
119 for (int i = 0; i <= n; i++)
120 for (int j = 0; j < ac.cnt; j++)
121 path[i][j] = "";
122 int ans = 0, num1 = 0, num2 = 0;
123 dp[0][0] = 0;
124 for (int i = 0; i < n; ++i) {
125 for (int j = 0; j < ac.cnt; ++j) {
126 if (dp[i][j] == -1) continue;
127 for (int k = 0; k < 26; ++k) {
128 int idx = ac.next[j][k];
129 if (dp[i + 1][idx] < dp[i][j] + ac.End[idx]) {
130 dp[i + 1][idx] = dp[i][j] + ac.End[idx];
131 path[i + 1][idx] = path[i][j] + char(k + 'a');
132 } else if (dp[i + 1][idx] == dp[i][j] + ac.End[idx] &&
133 path[i + 1][idx] > path[i][j] + char(k + 'a')) {
134 //fuck(path[i][j]);
135 path[i + 1][idx] = path[i][j] + char(k + 'a');
136 // fuck(path[i][j]);
137
138 }
139 }
140 }
141 }
142 for (int i = 1; i <= n; i++) {
143 for (int j = 0; j < ac.cnt; ++j) {
144 if (ans < dp[i][j]) {
145 ans = dp[i][j], num1 = i, num2 = j;
146 } else if (ans == dp[i][j]&&path[num1][num2] > path[i][j] && path[num1][num2].size()>=path[i][j].size()) {
147 num1 = i, num2 = j;
148 }
149 }
150 }
151 // printf("%d\n", ans);
152 cout << path[num1][num2] << endl;
153 }
154 return 0;
155 }