[洛谷P3975][TJOI2015]弦论

题目大意：求一个字符串的第$k$大字串，$t$表示长得一样位置不同的字串是否算多个

题解：$SAM$，先求出每个位置可以到达多少个字串（$Right$数组），然后在转移图上$DP$，若$t=1$，初始值赋成$Right$数组大小，否则赋成$1$

卡点：无

C++ Code：

#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#define maxn 500010

int n, t, k;
namespace SAM {
#define N (maxn << 1)
	int R[N], nxt[N][26], fail[N];
	int lst = 1, idx = 1, sz[N];
	void append(char __ch) {
		int ch = __ch - 'a';
		int p = lst, np = lst = ++idx; R[np] = R[p] + 1, sz[np] = 1;
		for (; p && !nxt[p][ch]; p = fail[p]) nxt[p][ch] = np;
		if (!p) fail[np] = 1;
		else {
			int q = nxt[p][ch];
			if (R[p] + 1 == R[q]) fail[np] = q;
			else {
				int nq = ++idx;
				fail[nq] = fail[q], R[nq] = R[p] + 1, fail[q] = fail[np] = nq;
				std::copy(nxt[q], nxt[q] + 26, nxt[nq]);
				for (; p && nxt[p][ch] == q; p = fail[p]) nxt[p][ch] = nq;
			}
		}
	}

	int sum[N];
	int buc[N], rnk[N];
	void make() {
		for (int i = 1; i <= idx; i++) ++buc[R[i]];
		for (int i = 1; i <= idx; i++) buc[i] += buc[i - 1];
		for (int i = idx; i; i--) rnk[buc[R[i]]--] = i;
		for (int i = idx; i; i--) {
			int u = rnk[i];
			sz[fail[u]] += sz[u];
			if (!t && u != 1) sz[u] = 1;
			sum[u] = sz[u];
			for (int j = 0; j < 26; j++) sum[u] += sum[nxt[u][j]];
		}
	}
	void print(int u) {
		if (k <= sz[u]) {
			putchar('\n');
			exit(0);
		}
		k -= sz[u];
		for (int i = 0; i < 26; i++) {
			int v = nxt[u][i];
			if (sum[v] >= k) putchar(i + 'a'), print(v);
			else k -= sum[v];
		}
	}
	void work() {
		make();
		if (sum[1] < k) {
			puts("-1");
			exit(0);
		}
		sz[1] = 0;
		print(1);
		putchar('\n');
	}
#undef N
}

char s[maxn];
int main() {
	scanf("%s", s); n = strlen(s);
	for (int i = 0; i < n; i++) SAM::append(s[i]);
	scanf("%d%d", &t, &k);
	SAM::work();
	return 0;
}