CF17E Palisection（回文自动机+链式前向星优化空间）

CF17E Palisection

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题意

给出字符串

求出字符串的子串中相交的对数

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思路

• 问题转化为字符串总对数减去不相交的回文串对数

  • 总对数为回文串总数：$n\ast (n-1)/2$

  • 不相交对数即为正反跑两边回文串

    对于以i结尾的回文串，与他不相交的回文串为反向的以x为结尾回文串总数$i\le x\le n$

• 处理二百万的字符串的Trie树明显会MLE，使用链式前向星维护Trie树

  • 即使用链式前向星创建字典树
  • 查询：遍历所有边，由于每个点在字典树访问次数有限，所以复杂不会超
  • 赋值：即为建边

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代码

链式前向星优化回文自动机

struct Edge{
	int v;
	int w;	//w为字母的值
	int next;
}edge[maxn];
int head[maxn], tot;
inline void AddEdge(int u, int v, int w) {
	edge[++tot].v = v;
	edge[tot].w = w;
	edge[tot].next = head[u];
	head[u] = tot;
}
int find(int now, int w) {	//寻找函数
	for (int i = head[now]; i; i = edge[i].next)
		if (edge[i].w == w) return edge[i].v;
	return 0;
}
for (int i = 1; i <= n; i++) {
	while (in[i - len[last] - 1] != in[i]) last = fail[last];
	int treenode = find(last, in[i] - 'a');//寻找节点
	if (!treenode) {
		len[++num] = len[last] + 2;
		int j = fail[last];
		while (in[i - len[j] - 1] != in[i]) j = fail[j];
		fail[num] = find(j, in[i] - 'a');//寻找节点
		AddEdge(last, num, in[i] - 'a');//建边
		mark[num] = mark[fail[num]] + 1;
		treenode = num;	//更新节点
	}
	last = treenode;
	cnt[i] = mark[last];
}

寻找不相交对数

scanf("%s", in + 1);in[0] = '#';
insert(in, n);		//正向回文自动机
for (int i = 1; i <= n; i++) pre_cnt[i] = cnt[i];//记录cnt[]
reverse(in + 1, in + 1 + n);//翻转
memset(head, 0, sizeof(head));	tot = 0;//初始化前向星
insert(in, n);//反向回文自动机
for (int i = 2; i <= n; i++)
    cnt[i] = (cnt[i - 1] + cnt[i]) % mod;//后缀和
LL ans = 0;
for (int i = 1; i <= n; i++)//统计相交对数
	ans = (ans + (LL(pre_cnt[i]) * cnt[n - i]) % mod) % mod;
printf("%I64d\n", ((LL(cnt[n]) * (cnt[n] - 1) / 2 % mod - ans) % mod + mod) % mod);//不相交对数

AC

#pragma comment(linker, "/STACK:102400000,102400000")
#pragma warning (disable:4996)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 2000005;
const LL mod = 51123987;
char in[maxn];
int len[maxn], fail[maxn];
int last, num;
int mark[maxn], cnt[maxn], pre_cnt[maxn];
struct Edge{
	int v;
	int w;
	int next;
}edge[maxn];
int head[maxn], tot;
inline void AddEdge(int u, int v, int w) {
	edge[++tot].v = v;
	edge[tot].w = w;
	edge[tot].next = head[u];
	head[u] = tot;
}
int find(int now, int w) {
	for (int i = head[now]; i; i = edge[i].next)
		if (edge[i].w == w) return edge[i].v;
	return 0;
}
void insert(char* in, int n) {
	len[1] = -1;
	fail[0] = 1;
	num = 1;
	last = 0;
	for (int i = 1; i <= n; i++) {
		while (in[i - len[last] - 1] != in[i]) last = fail[last];
		int treenode = find(last, in[i] - 'a');
		if (!treenode) {
			len[++num] = len[last] + 2;
			int j = fail[last];
			while (in[i - len[j] - 1] != in[i]) j = fail[j];
			fail[num] = find(j, in[i] - 'a');
			AddEdge(last, num, in[i] - 'a');
			mark[num] = mark[fail[num]] + 1;
			treenode = num;
		}
		last = treenode;
		cnt[i] = mark[last];
	}
}
int main() {
	int n;
	scanf("%d", &n);
	scanf("%s", in + 1);
	in[0] = '#';
	insert(in, n);
	for (int i = 1; i <= n; i++) pre_cnt[i] = cnt[i];
	reverse(in + 1, in + 1 + n);
	memset(head, 0, sizeof(head)); tot = 0;
	insert(in, n);
	for (int i = 2; i <= n; i++) cnt[i] = (cnt[i - 1] + cnt[i]) % mod;
	LL ans = 0;
	for (int i = 1; i <= n; i++)
		ans = (ans + (LL(pre_cnt[i]) * cnt[n - i]) % mod) % mod;
	printf("%I64d\n", ((LL(cnt[n]) * (cnt[n] - 1) / 2 % mod - ans) % mod + mod) % mod);
}