后缀数组解决在线的多模板匹配问题

终于学会倍增法了， 先一个最水最水的后缀数组应用。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e6;

char buf[maxn];
int str[maxn], len, sa[maxn];

inline int idx(char c) {
	return c - 'a';
}

int stra[maxn], strb[maxn], strcnt[maxn];

void build_sa(int *str, int *sa, int n, int m) {
	int i, j, *x = stra, *y = strb, *cnt = strcnt, chcnt;
	//第一次基数排序
	for (i = 0; i < m; i++) cnt[i] = 0;
	for (i = 0; i < n; i++) cnt[x[i] = str[i]]++; 
	for (i = 1; i < m; i++) cnt[i] += cnt[i - 1];
	for (i = n - 1; i >= 0; i--) sa[--cnt[x[i]]] = i;
	//倍增长度排序
	for (chcnt = 1, j = 1; chcnt < n; j <<= 1, m = chcnt) {
		//根据第二关键字排序，可以由上一次得到的sa值获得
		for (i = n - j, chcnt = 0; i < n; i++) y[chcnt++] = i;
		for (i = 0; i < n; i++) if (sa[i] >= j) y[chcnt++] = sa[i] - j;
		//根据第一关键字排序
		for (i = 0; i < m; i++) cnt[i] = 0;
		for (i = 0; i < n; i++) cnt[x[y[i]]]++;
		for (i = 1; i < m; i++) cnt[i] += cnt[i - 1];
		for (i = n - 1; i >= 0; i--) sa[--cnt[x[y[i]]]] = y[i];
		//根据sa值重新计算名次数组
		swap(x, y);
		for (chcnt = 1, x[sa[0]] = 0, i = 1; i < n; i++) {
			bool eql = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + j] == y[sa[i - 1] + j];
			x[sa[i]] = eql ? chcnt - 1 : chcnt++;
		}
	}
}

char dict[maxn];

void query(int len) {
	int l = 0, r = len, ansstr = 0, clen;
	clen = strlen(dict);
	while (l <= r) {
		int mid = (l + r) >> 1;
		bool ok = !strncmp(dict, buf + sa[mid], clen);
		if (ok) {
			ansstr = mid; r = mid - 1;
		}
		else l = mid + 1;
	}
	for (int i = ansstr; !strncmp(dict, buf + sa[i], clen); i++) {
		printf("find %s at pos %d
", dict, sa[i]);
	}
}


int main() {
	while (scanf("%s", buf) != EOF) {
		len = 0;
		while (buf[len] != 0) {
			str[len] = idx(buf[len]);
			len++;
		}
		str[len] = 0;
		build_sa(str, sa, len + 1, 27);
		int N; scanf("%d", &N);
		for (int i = 0; i < N; i++) {
			scanf("%s", dict);
			query(len + 1);
		}
	}
	return 0;
}