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UVA 10829 - L-Gap Substrings

题目链接

题意：一个字符串如果形如UGU，的形式，G的长度为L，那么称这个字符串为L串，给定一个字符串，问这个字符串子串为g串的个数

思路：做这题前先做了POJ3693，有一个思想就是枚举长度分段，这样的话对于一个U长度为l的而言，只要在当前位置和当前位置之后(l + g)的位置分别向前向后找lcp，两个lcp加起来的长度减去l就是可以可以的种数，累加起来就是答案

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

const int MAXLEN = 100005;

struct Suffix {

    int s[MAXLEN];
    int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;
    int rank[MAXLEN], height[MAXLEN];
    int best[MAXLEN][20];

    int g, len;
    char str[MAXLEN];

    void build_sa(int m) {
	n++;
	int i, *x = t, *y = t2;
	for (i = 0; i < m; i++) c[i] = 0;
	for (i = 0; i < n; i++) c[x[i] = s[i]]++;
	for (i = 1; i < m; i++) c[i] += c[i - 1];
	for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
	for (int k = 1; k <= n; k <<= 1) {
	    int p = 0;
	    for (i = n - k; i < n; i++) y[p++] = i;
	    for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;
	    for (i = 0; i < m; i++) c[i] = 0;
	    for (i = 0; i < n; i++) c[x[y[i]]]++;
	    for (i = 0; i < m; i++) c[i] += c[i - 1];
	    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
	    swap(x, y);
	    p = 1; x[sa[0]] = 0;
	    for (i = 1; i < n; i++)
		x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p++;
	    if (p >= n) break;
	    m = p;
	}
	n--;
    }

    void getHeight() {
	int i, j, k = 0;
	for (i = 1; i <= n; i++) rank[sa[i]] = i;
	for (i = 0; i < n; i++) {
	    if (k) k--;
	    int j = sa[rank[i] - 1];
	    while (s[i + k] == s[j + k]) k++;
	    height[rank[i]] = k;
	}
    }

    void initRMQ() {
	for (int i = 1; i <= n; i++) best[i][0] = height[i];
	for (int j = 1; (1<<j) <= n; j++)
	    for (int i = 1; i + (1<<j) - 1 <= n; i++)
		best[i][j] = min(best[i][j - 1], best[i + (1<<(j - 1))][j - 1]);
    }

    int lcp(int L, int R) {
	L = rank[L]; R = rank[R];
	if (L > R) swap(L, R);
	L++;
	int k = 0;
	while ((1<<(k + 1)) <= R - L + 1) k++;
	return min(best[L][k], best[R - (1<<k) + 1][k]);
    }

    void init() {
	n = 0;
	scanf("%d%s", &g, str);
	len = strlen(str);
	for (int i = 0; i < len; i++)
	    s[n++] = str[i] - 'a' + 1;
	s[n++] = 27;
	for (int i = len - 1; i >= 0; i--)
	    s[n++] = str[i] - 'a' + 1;
	s[n] = 0;
    }

    ll solve() {
	init();
	build_sa(28);
	getHeight();
	initRMQ();
	ll ans = 0;
	for (int d = 1; d < len / 2; d++) {
	    for (int j = 0; j < len; j += d) {
		int l = j, r = j + d + g, sum = 0;
		if (r < len) sum += min(lcp(l, r), d);
		if (l >= 1) sum += min(lcp(n - l, n - r), d - 1);
		ans += max(0, sum - d + 1);
	    }
	}
	return ans;
    }

} gao;

int t;

int main() {
    int cas = 0;
    scanf("%d", &t);
    while(t--) {
	printf("Case %d: %lld\n", ++cas, gao.solve());
    }
    return 0;
}

UVA 10829 - L-Gap Substrings(后缀数组)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38780843