codeforces gym 100548G

最新推荐文章于 2020-12-21 22:22:58 发布

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最新推荐文章于 2020-12-21 22:22:58 发布

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分类专栏： Codeforces ----回文树文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/jtjy568805874/article/details/51164757

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本文介绍了一种利用回文树的数据结构解决两个字符串间的公共回文子串计数问题的方法，并提供了三种不同的实现思路：双回文树DFS、合并构建单回文树及作为查询操作。

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求两个字符串的公共回文子串个数，搞两个回文树，dfs一遍即可。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
LL ans;
int T;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 2e5 + 10;
	const static int size = 26;
	int next[maxn][size], sz, tot;
	int fail[maxn], len[maxn], last;
	LL cnt[maxn];
	char s[maxn];
	LL operator[](const int &x) { return cnt[x]; }
	void clear()
	{
		len[1] = -1; len[2] = 0;
		fail[1] = fail[2] = 1;
		cnt[1] = cnt[2] = tot = 0;
		last = (sz = 3) - 1;	
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;  cnt[sz] = 0;  return sz++;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	int add(char pos)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last);
		if (next[y][x]) { last = next[y][x]; }
		else {
			last = next[y][x] = Node(len[y] + 2);
			fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
		}
		return ++cnt[last];
	}
	void work()
	{
		for (int i = sz - 1; i > 2; i--)
		{
			if (fail[i] > 2) cnt[fail[i]] += cnt[i];
		}
	}
}work[2];

void dfs(int x, int y)
{
	ans += work[0][x] * work[1][y];
	for (int i = 0; i < 26; i++)
	{
		if (work[0].next[x][i] && work[1].next[y][i])
		{
			dfs(work[0].next[x][i], work[1].next[y][i]);
		}
	}
}

int main()
{
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		for (int i = 0; i < 2; i++)
		{
			work[i].clear();
			scanf("%s", s);
			for (int j = 0; s[j]; j++)
			{
				work[i].add(s[j]);
			}
			work[i].work();
		}
		ans = 0;
		dfs(1, 1);
		dfs(2, 2);
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}

最近和同学讨论的时候又重新的想了这个问题，这里提供两个新的解决方案。

把第二个串也插进去一起统计的方法。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
LL ans;
int T;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 4e5 + 10;
	const static int size = 26;
	int next[maxn][size], sz, tot;
	int fail[maxn], len[maxn], last;
	LL cnt[maxn], t[maxn];
	char s[maxn];
	void clear()
	{
		len[1] = -1; len[2] = 0;
		fail[1] = fail[2] = 1;
		cnt[1] = cnt[2] = tot = 0;
		last = (sz = 3) - 2;
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;  t[sz] = cnt[sz] = 0;  return sz++;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	int add(char pos, int kind)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last);
		if (!(last = next[y][x]))
		{
			last = next[y][x] = Node(len[y] + 2);
			fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
		}
		return kind ? ++t[last] : ++cnt[last];
	}
	void work()
	{
		for (int i = sz - 1; i > 2; i--)
		{
			if (fail[i] > 2)
			{
				cnt[fail[i]] += cnt[i];
				t[fail[i]] += t[i];
			}
			ans += cnt[i] * t[i];
		}
	}
}work;

int main()
{
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		work.clear();
		scanf("%s", s);
		for (int i = 0; s[i]; i++) work.add(s[i], 0);
		scanf("%s", s);
		work.tot = 0;	work.last = 1;
		for (int i = 0; s[i]; i++) work.add(s[i], 1);
		ans = 0;
		work.work();
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}

不把第二个串插入，仅当做查询的方法，这里可以看出回文树和后缀自动机以及ac自动机有很多相同之处。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
LL ans;
int T;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 2e5 + 10;
	const static int size = 26;
	int next[maxn][size], sz, tot;
	int fail[maxn], len[maxn], last;
	LL cnt[maxn], t[maxn];
	char s[maxn];
	void clear()
	{
		len[1] = -1; len[2] = 0;
		fail[1] = fail[2] = 1;
		cnt[1] = cnt[2] = tot = 0;
		last = (sz = 3) - 2;
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;  t[sz] = cnt[sz] = 0;  return sz++;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	int add(char pos, int kind)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last);
		if (!(last = next[y][x]))
		{
			if (!kind)
			{
				last = next[y][x] = Node(len[y] + 2);
				fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
			}
			else
			{
				while (!next[y][x] && y > 1) y = getfail(fail[y]);
				if (!(last = next[y][x])) last = 1;
			}
		}
		return kind ? ++t[last] : ++cnt[last];
	}
	void work()
	{
		for (int i = sz - 1; i > 2; i--)
		{
			if (fail[i] > 2)
			{
				cnt[fail[i]] += cnt[i];
				t[fail[i]] += t[i];
			}
			ans += cnt[i] * t[i];
		}
	}
}work;

int main()
{
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		work.clear();
		scanf("%s", s);
		for (int i = 0; s[i]; i++) work.add(s[i], 0);
		scanf("%s", s);
		work.tot = 0;	work.last = 1;
		for (int i = 0; s[i]; i++) work.add(s[i], 1);
		ans = 0;
		work.work();
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}