给定
第一眼没什么想法,看了一眼数据范围才反应过来这似乎就是个暴力,我们尝试分析一下:
令
化简一下:
这一大坨东西我不知道我是否分析错了,不过大概也八九不离十,总之这个东西我们随便带入几个
做法的话先通过调用 next_permutation()
函数枚举全排列,此时还会多一个
然后每次跑一遍深搜,用 string
的本质是 basic_string
,且 basic_string
支持 +
和 +=
等的特性,即可十分便捷地实现一般的深搜思路,注意每两个之间至少插入一个下划线,且注意需要限制最多额外添加
xxxxxxxxxx
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11using namespace std;
12
13mt19937 rnd(random_device{}());
14int rndd(int l, int r){return rnd() % (r - l + 1) + l;}
15bool rnddd(int x){return rndd(1, 100) <= x;}
16
17typedef unsigned int uint;
18typedef unsigned long long unll;
19typedef long long ll;
20typedef long double ld;
21
22template < typename T = int >
23inline T read(void);
24
25int N, M;
26int mx(-1);
27string S[10];
28unordered_set < string > block;
29string ans;
30
31void dfs(int dep = 1, int lft = mx, string cur = string()){
32 if(dep > N){
33 if(3 <= (int)cur.length() && (int)cur.length() <= 16 && block.find(cur) == block.end())ans = cur;
34 return;
35 }
36 if(dep == 1)return dfs(dep + 1, lft, cur + S[dep]);
37 dfs(dep + 1, lft, cur + "_" + S[dep]);
38 for(int i = 1; i <= lft; ++i)
39 cur += "_", dfs(dep + 1, lft - i, cur + "_" + S[dep]);
40}
41
42int main(){
43 N = read(), M = read();
44 int slen(0);
45 for(int i = 1; i <= N; ++i)cin >> S[i], slen += S[i].length();
46 for(int i = 1; i <= M; ++i){string T; cin >> T; block.insert(T);}
47 mx = 16 - slen - N + 1;
48 if(mx < 0)printf("-1\n"), exit(0);
49 int tot(1); for(int i = 1; i <= N; ++i)tot *= i;
50 for(int i = 1; i <= tot; ++i)dfs(), next_permutation(S + 1, S + N + 1);
51 if(ans.empty())printf("-1\n"), exit(0);
52 cout << ans << endl;
53 fprintf(stderr, "Time: %.6lf\n", (double)clock() / CLOCKS_PER_SEC);
54 return 0;
55}
56
57template < typename T >
58inline T read(void){
59 T ret(0);
60 int flag(1);
61 char c = getchar();
62 while(c != '-' && !isdigit(c))c = getchar();
63 if(c == '-')flag = -1, c = getchar();
64 while(isdigit(c)){
65 ret *= 10;
66 ret += int(c - '0');
67 c = getchar();
68 }
69 ret *= flag;
70 return ret;
71}
update-2023_01_18 初稿