题目大意：给出一个字符串，给出一些询问，每次问几个后缀两两之间的LCP之和。

思路：保证Σask数量级在O(n)上，可以考虑一下虚树了。建立虚树之后，这题就和差异那个题一样了。但是必须要打时间戳啊，要不死的很惨的啊。。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1000010
using namespace std;

struct Complex{
	Complex *tranc[26],*father;
	int len,id;
}none,*nil = &none,mempool[MAX],*C = mempool,*root,*last;
int cnt = -1;

Complex *NewComplex(int _)
{
	C->id = ++cnt;
	C->len = _;
	fill(C->tranc,C->tranc + 26,nil);
	C->father = nil;
	return C++;
}

void Initialize()
{
	root = last = NewComplex(0);
	fill(nil->tranc,nil->tranc + 26,nil);
	nil->father = nil;
}

int length,asks;
char s[MAX];

inline void Add(int c)
{
	Complex *np = NewComplex(last->len + 1),*p = last;
	for(; p != nil && p->tranc[c] == nil; p = p->father)
		p->tranc[c] = np;
	if(p == nil)	np->father = root;
	else {
		Complex *q = p->tranc[c];
		if(q->len == p->len + 1)	np->father = q;
		else {
			Complex *nq = NewComplex(p->len + 1);
			memcpy(nq->tranc,q->tranc,sizeof(q->tranc));
			nq->father = q->father;
			np->father = q->father = nq;
			for(; p != nil && p->tranc[c] == q; p = p->father)
				p->tranc[c] = nq;	
		}
	}
	last = np;
}

int head[MAX],total;
int next[MAX],aim[MAX];

int deep[MAX],father[MAX][20];

void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

int pos[MAX];

void DFS(int x,int last)
{
	static int c = 0;
	deep[x] = deep[last] + 1;
	father[x][0] = last;
	pos[x] = ++c;
	for(int i = head[x]; i; i = next[i])
		DFS(aim[i],x);
}

void SparseTable()
{
	for(int j = 1; j <= 19; ++j)
		for(int i = 1; i <= cnt; ++i)
			father[i][j] = father[father[i][j - 1]][j - 1];
}

inline int GetLCA(int x,int y)
{
	if(deep[x] < deep[y])	swap(x,y);
	for(int i = 19; ~i; --i)
		if(deep[father[x][i]] >= deep[y])
			x = father[x][i];
	if(x == y)	return x;
	for(int i = 19; ~i; --i)
		if(father[x][i] != father[y][i])
			x = father[x][i],y = father[y][i];
	return father[x][0];
}

bool cmp(int x,int y)
{
	return pos[x] < pos[y];
}

int src[MAX];

namespace Graph{
	int head[MAX],v_head[MAX],total;
	int next[MAX],aim[MAX];
	int father[MAX],v_father[MAX];

	int T;
	int ans;
	int size[MAX];
	int v_set[MAX];
	bool set[MAX];

	void Reset() {
		++T;
		total = 0;
		ans = 0;
	}
	void Add(int x,int y) {
		if(v_head[x] != T) {
			v_head[x] = T;
			head[x] = 0;
		}
		next[++total] = head[x];
		aim[total] = y;
		head[x] = total;
	}
	void TreeDP(int x,int last) {
		if(v_set[x] != T)	v_set[x] = T,set[x] = false;
		size[x] = set[x];
		if(v_head[x] != T) {
			v_head[x] = T;
			head[x] = 0;
		}
		for(int i = head[x]; i; i = next[i]) {
			TreeDP(aim[i],x);
			size[x] += size[aim[i]];
		}
		ans += size[x] * (size[x] - 1) * (mempool[x].len - mempool[last].len) >> 1;
	}
}

int p[MAX];

int main()
{
	cin >> length >> asks;
	scanf("%s",s);
	Initialize();
	for(int i = strlen(s) - 1; ~i; --i)
		p[i + 1] = cnt + 1,Add(s[i] - 'a');
	for(int i = 1; i <= cnt; ++i)
		Add(mempool[i].father->id,mempool[i].id);
	DFS(0,0);
	SparseTable();
	while(asks--) {
		int num;
		scanf("%d",&num);
		static int v[MAX],T = 0;
		++T;
		int temp = 0;
		for(int x,i = 1; i <= num; ++i) {
			scanf("%d",&x);
			if(v[x] != T)	
				src[++temp] = p[x],v[x] = T;
		}
		num = temp;
		sort(src + 1,src + num + 1,cmp);
		Graph::Reset();
		int top = 0,end = num;
		static int stack[MAX];
		for(int i = 1; i <= num; ++i) {
			if(!top) {
				stack[++top] = src[i];
				Graph::v_set[src[i]] = Graph::T;
				Graph::set[src[i]] = true;
				continue;
			}
			int lca = GetLCA(stack[top],src[i]);
			Graph::v_father[src[i]] = Graph::T;
			Graph::father[src[i]] = lca;
			while(deep[stack[top]] > deep[lca]) {
				if(deep[stack[top - 1]] <= deep[lca]) {
					Graph::v_father[stack[top]] = Graph::T;
					Graph::father[stack[top]] = lca;
				}
				--top;
			}
			if(stack[top] != lca) {
				Graph::v_father[lca] = Graph::T;
				Graph::father[lca] = stack[top],stack[++top] = src[++end] = lca;
			}
			stack[++top] = src[i];
			Graph::v_set[src[i]] = Graph::T;
			Graph::set[src[i]] = true;
		}
		for(int i = 1; i <= end; ++i)
			if(Graph::v_father[src[i]] == Graph::T)
				Graph::Add(Graph::father[src[i]],src[i]);
		if(Graph::v_head[0] != T) {
			Graph::v_head[0] = T;
			Graph::head[0] = 0;
		}
		Graph::TreeDP(0,0);
		printf("%d\n",Graph::ans);
	}
	return 0;
}

BZOJ 3879 SvT 后缀树+虚树