HDU 2144 Evolution 字尾樹/字尾陣列

題意

• 給我們不到一百個字串（長度不到一百）以及一個百分比q，然後如果某兩個字串的最長公共子串佔比超過了q（在兩個串中都超過）則兩個串為一個集合，問最終能分出幾個集合

思路

• 顯然我們列舉每兩個字串是n2複雜度，那麼判斷最長公共子串必須要n的複雜度才能過這道題。

• 所以使用字尾陣列或者字尾樹，每次都重建一遍，儘管常數可能不太優秀但是n3複雜度還是可以過的。這次把超過理解成了大於等於，所以wa了兩次，可惜，程式碼中字尾樹用的是之前的模板。

AC程式碼

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>

using namespace std;

const int maxn = (1 << 30);
const int root = 1;

char ss[210] = {0};
char ss2[205] = {0};
int act = 1, co = 1;
int acteg = -1;
int tep = 0;
int ind = 0, rem = 0, s_end = -1;
int links[205] = {0};
int vv[205] = {0};
int mm = 0;
int linkk = 0;
int len1 = 0, len2 = 0;

int ans = 0;

struct ab
{
	int l;
	int r;
	int nex;
	int alp[28];
} tree[1005];		// 作為分割與結束符 (ascii相鄰防止越界)

char sss[105][105];

int add_new(int o, int ll = s_end, int rr = maxn)
{
	tree[o].l = ll;
	tree[o].r = rr;
	return o;
}

void add_link(int o)
{
	if (linkk)
	{
		tree[linkk].nex = o;
	}
	linkk = o;
}

int check_len(int o)
{
	return min(tree[o].r, s_end) - tree[o].l + 1;
}

bool check_contain(int o)
{
	int node_len = check_len(o);
	if (node_len <= ind)
	{
		ind -= node_len;
		tep += node_len;
		act = o;
		return true;
	}
	return false;
}

void add(char cc)
{
	++rem;
	linkk = 0;
	while (rem > 0)
	{
		if (!ind)
		{
			tep = s_end;
		}
		int& actedge_node = tree[act].alp[ss[tep] - 'A'];
		if (!actedge_node)
		{
			actedge_node = add_new(++co, s_end);
			add_link(act);
		}
		else
		{
			if (check_contain(actedge_node))
			{
				continue;
			}
			else
			{
				if (ss[tree[actedge_node].l + ind] != cc)	// 分裂注意原樹(actedge_node)必須成為子樹（否則會和原先的子樹失去聯絡） 
				{
					int leaf1 = add_new(++co, s_end);
					int leaf2 = actedge_node;
					int newtree = add_new(++co, tree[actedge_node].l, tree[actedge_node].l + ind - 1);
					tree[newtree].alp[cc - 'A'] = leaf1;
					tree[newtree].alp[ss[tree[actedge_node].l + ind] - 'A'] = leaf2;
					tree[leaf2].l += ind;
					actedge_node = newtree;
					add_link(actedge_node);
				}
				else
				{
					++ind;		// 活躍半徑只在此處增加 ，增加完就加鏈並結束本次增點 
//					if (act != root)
//					{
						add_link(act);
//					}
					break;
				}
			}
		}
		--rem;
		if (act == root)
		{
			if (!ind)
			{
				break;
			}
			tep = s_end - rem + 1;
			--ind;
		}
		else
		{
//			ind = rem - 1;
//			tep = s_end - rem + 1;
			if (tree[act].nex)
			{
				act = tree[act].nex;
			}
			else
			{
				act = root;
			}
		}
	}
}

int dfs(int o, int cc)		// 本題所需的搜尋 返回1代表包含{，2代表包含|，3代表都有 
{
	bool bk1 = false;
	bool bk2 = false;
	bool stop = false;
	for (int i = 0; i <= 27; ++i)
	{
		if (tree[o].alp[i])
		{
			if (tree[tree[o].alp[i]].r != maxn)
			{
				int contain_terminal = dfs(tree[o].alp[i], cc + check_len(tree[o].alp[i]));
				if (contain_terminal == 1)
				{
					bk1 = true;
				}
				if (contain_terminal == 2)
				{
					bk2 = true;
				}
				if (contain_terminal == 3)
				{
					bk1 = bk2 = true;
					stop = true;
				}
			}
			else
			{
				if (tree[tree[o].alp[i]].l > len1)
				{
					bk2 = true;
				}
				else
				{
					bk1 = true;
				}
			}
		}
	}
	if (stop)
	{
		return 3;
	}
	if (bk1 && bk2)
	{
		ans = max(ans, cc);
		return 3;
	}
	if (bk1)
	{
		return 1;
	}
	if (bk2)
	{
		return 2;
	}
}

int suffixt()
{
	len1 = strlen(ss);
	len2 = strlen(ss2);
	
	memset(links, 0, sizeof(links));
	memset(tree, 0, sizeof(tree));
	memset(vv, 0, sizeof(vv));

	act = 1, co = 1;
	acteg = -1;
	tep = 0;
	ind = 0, rem = 0, s_end = -1;
	mm = 0;
    linkk = 0;
	ans = 0;
	
	ss[len1] = 'Z' + 1;					//ss1的結束符，防止兩字串字尾拼接  
	for (int i = len1 + 1; i <= len1 + len2; ++i)
	{
		ss[i] = ss2[i - len1 - 1];
	}
	ss[len1 + len2 + 1] = 'Z' + 2;		//ss2的結束符（也是整個合串的結束符） 
	for (int i = 0; i <= len1 + len2 + 1; ++i)
	{
		++s_end;
		add(ss[i]);
	}
	dfs(root, 0);
	return ans;
	return 0;
}

int ff[105] = {0};
int f(int x)
{
	if (ff[x] != x)
	{
		return ff[x] = f(ff[x]);
	}
	else
	{
		return x;
	}
}

int main()
{
	int n;
    double q;
    int kk = 0;
	while (scanf("%d%lf", &n, &q) == 2)
	{
		q /= 100;
		for (int i = 1; i <= n; ++i)
		{
			ff[i] = i;
		}
		for (int i = 1; i <= n; ++i)
		{
			scanf("%s", sss[i]);
			for (int j = 1; j < i; ++j)
			{
				int a = f(i), b = f(j);
				if (a == b)
				{
					continue;
				}
				int k;
				for (k = 0; sss[j][k]; ++k)
				{
					ss2[k] = sss[j][k];
				}
				int len11 = k;
				ss2[k] = 0;
				for (k = 0; sss[i][k]; ++k)
				{
					ss[k] = sss[i][k];
				}
				len11 = max(len11, k);
				ss[k] = 0;
				int cc = suffixt();
				if ((double)cc / (double)len11 > q)
				{
					ff[a] = b;
				}
			}
		}
		int bkk[105], anss = 0;
		memset(bkk, 0, sizeof(bkk));
		for (int i = 1; i <= n; ++i)
		{
			int c = f(i);
			if (!bkk[c])
			{
				++anss;
				bkk[c] = true;
			}
		}
		printf("Case %d:\n%d\n", ++kk, anss);
	}
	return 0;
}