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于 2020-11-04 10:02:15 发布

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本文介绍了一种使用后缀数组解决特定问题的方法，并针对该方法进行了优化。通过将问题转化为求所有height之和，首先采用暴力解法获得初步答案。接着，通过对插入新字符导致后缀排名变化的现象进行分析，提出利用前缀而非后缀来保持height之和不变的策略，进一步通过set记录前驱后继，结合字符串反转简化前缀排序。

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答案是所有height之和
很容易想到直接用后缀数组暴力骗60分

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define FOR(i, x, y) for (int i = x; i <= y; i++)
#define DOW(i, x, y) for (int i = x; i >= y; i--)
using namespace std;

const int N = 1e5 + 10;

int a[N], b[N], s[N];
int x[N << 1], y[N << 1], c[N], sa[N], rk[N], height[N];

bool cmp(int i, int k) {
	return y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k];
}

void get_sa(int n, int m) {
	FOR(i, 1, m) c[i] = 0;
	FOR(i, 1, n) c[x[i] = s[i]]++;
	FOR(i, 2, m) c[i] += c[i - 1];
	DOW(i, n, 1) sa[c[x[i]]--] = i;
	for (int k = 1; k <= n; k <<= 1) {
		int z = 0;
		FOR(i, n - k + 1, n) y[++z] = i;
		FOR(i, 1, n) if (sa[i] > k) y[++z] = sa[i] - k;
		FOR(i, 1, m) c[i] = 0;
		FOR(i, 1, n) c[x[i]]++;
		FOR(i, 2, m) c[i] += c[i - 1];
		DOW(i, n, 1) sa[c[x[y[i]]]--] = y[i], y[i] = 0;
		swap(x, y); x[sa[1]] = m = 1;
		for (int i = 2; i <= n; i++)
			x[sa[i]] = cmp(i, k) ? m : ++m;
		if (n == m) break;
	}
}

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

int main() {
	int n, m;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		b[i] = a[i];
	}
	sort(b + 1, b + n + 1);
	m = unique(b + 1, b + n + 1) - b - 1;
	for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
	memcpy(s, a, sizeof(s));
	for (int i = 1; i <= n; i++) {
		get_sa(i, m);
		get_height(i);
		int ans = (i + 1) * i / 2;
		for (int j = 2; j <= n; j++) ans -= height[j];
		printf("%d\n", ans);
	}
	return 0;
}

然后考虑优化。
题目说的是每一次新插入一个字符，这就会导致后缀的排名变化，怎么办呢？
注意，插入一个会让后缀的排名变化，但如果是前缀呢？
用前缀搞出来的height之和是不会变化的。
而每一次只需要插入一个新的前缀，先处理好以后用set记录前驱后继即可。
对于前缀的排序，也不需要吧后缀数组硬改成“前缀”数组，把字符串反过来就可以了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <set>
#define FOR(i, x, y) for (int i = x; i <= y; i++)
#define DOW(i, x, y) for (int i = x; i >= y; i--)
using namespace std;

const int N = 1e5 + 10;

int a[N], b[N];
int x[N << 1], y[N << 1], c[N], sa[N], rk[N], height[N];
int f[N][21], Log[N];

bool cmp(int i, int k) {
	return y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k];
}

void get_sa(int n, int m) {
	FOR(i, 1, m) c[i] = 0;
	FOR(i, 1, n) c[x[i] = a[i]]++;
	FOR(i, 2, m) c[i] += c[i - 1];
	DOW(i, n, 1) sa[c[x[i]]--] = i;
	for (int k = 1; k <= n; k <<= 1) {
		int z = 0;
		FOR(i, n - k + 1, n) y[++z] = i;
		FOR(i, 1, n) if (sa[i] > k) y[++z] = sa[i] - k;
		FOR(i, 1, m) c[i] = 0;
		FOR(i, 1, n) c[x[i]]++;
		FOR(i, 2, m) c[i] += c[i - 1];
		DOW(i, n, 1) sa[c[x[y[i]]]--] = y[i], y[i] = 0;
		swap(x, y); x[sa[1]] = m = 1;
		for (int i = 2; i <= n; i++)
			x[sa[i]] = cmp(i, k) ? m : ++m;
		if (n == m) break;
	}
}

void get_height(int n) {
	for (int i = 1; i <= n; i++) rk[sa[i]] = i;
	for (int i = 1, j, k = 0; i <= n; i++) {
		j = sa[rk[i] - 1];
		if (k) k--;
		while (i + k <= n && j + k <= n && a[i + k] == a[j + k]) k++;
		height[rk[i]] = k;
	}
}

void get_st(int n) {
	int t = log(n) / log(2);
	Log[0] = -1;
	for (int i = 1; i <= n; i++) {
		Log[i] = Log[i >> 1] + 1;
		f[i][0] = height[i];
	}
	for (int j = 1; j <= t; j++)
		for (int i = 1; i + (1 << j) - 1 <= n; i++)
			f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}

int calc(int l, int r) {
	int t = Log[r - l + 1];
	return min(f[l][t], f[r - (1 << t) + 1][t]);
}

int main() {
	int n, m;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		b[i] = a[i];
	}
	sort(b + 1, b + n + 1);
	m = unique(b + 1, b + n + 1) - b - 1;
	for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
	reverse(a + 1, a + n + 1);
	get_sa(n, m), get_height(n), get_st(n);
	set<int> s;
	set<int>::iterator it;
	long long ans = 0;
	for (int i = n; i; i--) {
		s.insert(rk[i]);
		it = s.find(rk[i]);
		int k = 0;
		if (it != s.begin()) {
			--it;
			k = calc((*it) + 1, rk[i]);
			it++;
		}
		it++; 
		if (it != s.end()) 
			k = max(k, calc(rk[i] + 1, *it)); 
		//取max是因为当前的前缀把其前驱后继之间分成两段，记a,b
		//ans已经记录min(a,b)，要求a+b，只用加上max(a,b) 
		ans += (long long)(n + 1 - i - k);
		printf("%lld\n", ans);
	}
	return 0;
}