【AHOI2013】差异

题面

题解

$ \because \sum_{1 \leq i < j \leq n} i + j = \frac{n(n-1)(n+1)}2 $

所以只需求$\sum lcp(i,j)$即可。

$ \because lcp(i,j)=\min_{rank[i] \leq k \leq rank[j]}\{height[k]\} $

所以可以选用最小值分治算法：

int min[maxn];
long long ans;
void Div(int l, int r)
{
    if(l == r) return (void)(ans += height[l]);
    int mid = (l + r) >> 1;
    Div(l, mid); Div(mid + 1, r);
    min[mid] = height[mid];
    min[mid + 1] = height[mid + 1];
    for(RG int i = mid - 1; i >= l; i--)
        min[i] = std::min(min[i + 1], height[i]);
    for(RG int i = mid + 2; i <= r; i++)
        min[i] = std::min(min[i - 1], height[i]);
    int j = mid;
    for(RG int i = mid; i >= l; i--)
    {
        while(j < r && min[j + 1] >= min[i]) ++j;
        ans += 1ll * min[i] * (j - mid);
    }
    j = mid + 1;
    for(RG int i = mid + 1; i <= r; i++)
    {
        while(j > l && min[j - 1] > min[i]) --j;
        ans += 1ll * min[i] * (mid + 1 - j);
    }
}
//...
Div(1, n);

但是我们要精益求精，我们可以想一想$O(n)$的算法。

用栈维护前面与$i$最近且小于等于$height[i]$的元素$p$

则转移方程为：

$ f[i]=f[p]+(i-p)\times height[i] $

//...
long long f[maxn];
struct node { int val, pos; };
std::stack<node> stk;

int main()
{
    //...
	long long ans = 0; int pos = 0;
	for(RG int i = 1; i <= n; i++)
	{
		int p = pos;
		while(!stk.empty() && stk.top().val > height[i]) stk.pop();
		if(!stk.empty()) p = stk.top().pos;
		ans += (f[i] = f[p] + (i - p) * height[i]);
		if(!height[i]) pos = i;
		stk.push((node){height[i], i});
	}
}

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<stack>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(500010);
char s[maxn];
int n, sa[maxn], rank[maxn], height[maxn];

void sort(int m)
{
	static int t[maxn], t2[maxn], c[maxn];
	int i, *x = t, *y = t2, p = 0;
	std::fill(c + 1, c + m + 1, 0);
	for(i = 1; i <= n; i++) ++c[x[i] = s[i]];
	for(i = 1; i <= m; i++) c[i] += c[i - 1];
	for(i = n; i; i--) sa[c[x[i]]--] = i;
	for(RG int k = 1; k <= n && p < n; k <<= 1)
	{
		p = 0;
		for(i = n - k + 1; i <= n; i++) y[++p] = i;
		for(i = 1; i <= n; i++) if(sa[i] > k) y[++p] = sa[i] - k;
		std::fill(c + 1, c + m + 1, 0);
		for(i = 1; i <= n; i++) ++c[x[y[i]]];
		for(i = 1; i <= m; i++) c[i] += c[i - 1];
		for(i = n; i; i--) sa[c[x[y[i]]]--] = y[i];
		std::swap(x, y), p = 1, x[sa[1]] = 1;
		for(i = 2; i <= n; i++)
			x[sa[i]] = (y[sa[i]] == y[sa[i - 1]]
			&& y[sa[i] + k] == y[sa[i - 1] + k]) ? p : ++p;
		m = p;
	}
}

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

long long f[maxn];
struct node { int val, pos; };
std::stack<node> stk;

int main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	scanf("%s", s + 1); n = strlen(s + 1);
	sort(130); get_height();
	long long ans = 0; int pos = 0;
	for(RG int i = 1; i <= n; i++)
	{
		int p = pos;
		while(!stk.empty() && stk.top().val > height[i]) stk.pop();
		if(!stk.empty()) p = stk.top().pos;
		ans += (f[i] = f[p] + (i - p) * height[i]);
		if(!height[i]) pos = i;
		stk.push((node){height[i], i});
	}
	printf("%lld\n", 1ll * n * (n - 1) * (n + 1) / 2 - ans * 2);
	return 0;
}