bzoj1195 [HNOI2006]最短母串 AC 自动机+状压+bfs

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=1195

题解

建立 AC 自动机,然后构建出 trie 图。

然后直接在 trie 图上走。但是我们需要经过每一个串。

所以我们处理一下每个点代表了哪些串,然后把状态加入进 bfs 状态。

然后 bfs 就可以了。


\(O(n)\)

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 12 + 1;
const int M = 50 + 1;
const int NM = 12 * 50 + 1;
const int NP = (1 << 12);
const int NMP = 12 * 50 * ((1 << 12) - 1) + 1;

int n, m, nod;
int q[NMP], q2[NMP], fr[NM][NP];
char s[N], ans[NM];
struct Node { int c[26], fa, v; char s; } t[NM];

inline void ins(char *s, int id) {
	int n = strlen(s + 1), o = 0;
	for (int i = 1; i <= n; ++i) {
		if (!t[o].c[s[i] - 'A']) t[o].c[s[i] - 'A'] = ++nod, t[nod].s = s[i];
		o = t[o].c[s[i] - 'A'];
	}
	t[o].v |= 1 << (id - 1);
}
inline void build() {
	int hd = 0, tl = 0;
	for (int i = 0; i < 26; ++i) if (t[0].c[i]) q[++tl] = t[0].c[i];
	while (hd < tl) {
		int x = q[++hd]; t[x].v |= t[t[x].fa].v;
		for (int i = 0; i < 26; ++i)
			if (t[x].c[i]) t[t[x].c[i]].fa = t[t[x].fa].c[i], q[++tl] = t[x].c[i];
			else t[x].c[i] = t[t[x].fa].c[i];
	}
}

inline void work() {
	int hd = 0, tl = 1, S = (1 << n) - 1;
	q[tl] = 0, q2[tl] = 0, fr[0][0] = -1;
	while (hd < tl) {
		int x = q[++hd];
		int s = q2[hd];
//		dbg("x = %d, s = %d\n", x, s);
		for (int i = 0; i < 26; ++i) if (t[x].c[i]) {
			int y = t[x].c[i], ss = s | t[y].v;
//			dbg("y = %d, ss = %d\n", y, ss);
			if (!fr[y][ss]) {
				fr[y][ss] = hd, q[++tl] = y, q2[tl] = ss;
//				dbg("****** y = %d, ss = %d\n", y, ss);
				if (ss == S) {
					int len = 0, z = y, zs = ss, zz;
					while (~fr[z][zs]) ans[++len] = t[z].s, zz = fr[z][zs], z = q[zz], zs = q2[zz];
					while (len) putchar(ans[len--]);
					puts("");
//					dbg("*********\n");
					return;
				}
				assert(fr[y][ss]);
			}
		}
	}
}

inline void init() {
	read(n);
	for (int i = 1; i <= n; ++i) scanf("%s", s + 1), ins(s, i);
	build();
//	dbg("*************\n");
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}
posted @ 2019-11-02 22:20  hankeke  阅读(160)  评论(0编辑  收藏  举报