BZOJ3238 [Ahoi2013]差异 【后缀数组 + 单调栈】

题目链接

BZOJ3238

题解

简单题
经典后缀数组 + 单调栈套路，求所有后缀(lcp)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define cls(s) memset(s,0,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
char s[maxn];
int sa[maxn],rank[maxn],height[maxn],bac[maxn],t1[maxn],t2[maxn],n,m;
void getsa(){
	int *x = t1,*y = t2; m = 255;
	for (int i = 0; i <= m; i++) bac[i] = 0;
	for (int i = 1; i <= n; i++) bac[x[i] = s[i]]++;
	for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
	for (int i = n; i; i--) sa[bac[x[i]]--] = i;
	for (int k = 1; k <= n; k <<= 1){
		int p = 0;
		for (int i = n - k + 1; i <= n; i++) y[++p] = i;
		for (int i = 1; i <= n; i++) if (sa[i] - k > 0) y[++p] = sa[i] - k;
		for (int i = 0; i <= m; i++) bac[i] = 0;
		for (int i = 1; i <= n; i++) bac[x[y[i]]]++;
		for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
		for (int i = n; i; i--) sa[bac[x[y[i]]]--] = y[i];
		swap(x,y);
		x[sa[1]] = p = 1;
		for (int 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);
		if (p >= n) break;
		m = p;
	}
	for (int i = 1; i <= n; i++) rank[sa[i]] = i;
	for (int i = 1,k = 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;
	}
}
cp st[maxn],t;
int top;
void solve(){
	LL sum = 0,ans = 0;
	for (int i = 2; i <= n; i++){
		t = mp(height[i],1);
		while (top && st[top].first >= t.first){
			sum -= 1ll * st[top].first * st[top].second;
			t.second += st[top].second;
			top--;
		}
		st[++top] = t;
		sum += 1ll * t.first * t.second;
		ans += sum;
	}
	printf("%lld
",1ll * n * (n + 1) * (n - 1) / 2 - 2 * ans);
}
int main(){
	scanf("%s",s + 1); n = strlen(s + 1);
	getsa();
	solve();
	return 0;
}