【BZOJ2806】【CTSC2012】Cheat 广义后缀自动机+二分+Dp

题目

题目在这里

思路&做法

我们先对标准作文库建广义后缀自动机。

然后对于每一篇阿米巴的作文, 我们首先把放到广义后缀自动机跑一遍, 对于每一个位置, 记录公共子串的长度\((\)即代码和下文中的\(val\)数组\()\)

接着我们二分答案, 用DP检验。
Dp方程很好想, \(d_i = max \{ d_j + i - j \ | \ i-val_i <= j <= i-lim \}\)

可以用单点队列优化。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

const int N = 1300010; //数组不能太大, 会T的

int n, m;

int val[N];

struct Suffix_Automaton
{	int nxt[N][2], fail[N], sz;
	int len[N];
	int root;

	Suffix_Automaton() { }

	inline int newnode(int l)
	{	memset(nxt[sz], 0, sizeof(nxt[sz]));
		fail[sz] = 0;
		len[sz] = l;
		return sz++;
	}

	void init()
	{	sz = 1;
		root = newnode(0);
	}

	inline int idx(char x) { return x - '0'; }

	int add(int last, char x)
	{	int c = idx(x);
		if (nxt[last][c])
		{	int p = last, q = nxt[last][c];
			if (len[q] == len[p] + 1)
				return q;
			else
			{	int u = newnode(len[p] + 1);
				for (int i = 0; i < 2; i++) nxt[u][i] = nxt[q][i];
				fail[u] = fail[q];
				fail[q] = u;
				while (p && nxt[p][c] == q)
				{	nxt[p][c] = u;
					p = fail[p];
				}
				return u;
			}
		}
		else
		{	int now = newnode(len[last] + 1);
			int p = last;
			while (p && !nxt[p][c])
			{	nxt[p][c] = now;
				p = fail[p];
			}
			if (!p) fail[now] = root;
			else
			{	int q = nxt[p][c];
				if (len[q] == len[p] + 1)
					fail[now] = q;
				else
				{	int u = newnode(len[p] + 1);
					for (int i = 0; i < 2; i++) nxt[u][i] = nxt[q][i];
					fail[u] = fail[q];
					fail[now] = fail[q] = u;
					while (p && nxt[p][c] == q)
					{	nxt[p][c] = u;
						p = fail[p];
					}
				}
			}
			return now;
		}
	}

	void insert(char *s)
	{	int Len = strlen(s);
		int last = root;
		for (int i = 0; i < Len; i++)
			last = add(last, s[i]);
	}

	void work(char *str)
	{	int cnt = 0;
		int now = root;
		int Len = strlen(str+1);
		for (int i = 1; i <= Len; i++)
		{	int c = idx(str[i]);
			if (nxt[now][c])
			{	cnt++;
				now = nxt[now][c];
			}
			else
			{	while (now && !nxt[now][c]) now = fail[now];
				if (!now) { now = root; cnt = 0; }
				else { cnt = len[now] + 1; now = nxt[now][c]; }
			}
			val[i] = cnt;
		}
	}
} tzw;

int d[N];
int Q[N], hd, tl;
bool check(char *s, int lim)
{	int Len = strlen(s+1);
	for (int i = 0; i <= lim; i++) d[i] = 0;
	hd = 0, tl = 1;
	for (register int i = lim; i <= Len; i++)
	{	d[i] = d[i-1];
		if (i > lim) //这里一定不能去掉, 去掉会RE
		{	while (hd < tl && Q[hd] < i-val[i]) hd++;
			while (hd < tl && d[Q[tl-1]]+i-Q[tl-1] < d[i-lim]+i-(i-lim)) tl--;
			Q[tl++] = i - lim;
		}
		if (val[i] >= lim) d[i] = max(d[i], d[Q[hd]] + i - Q[hd]);
	}
	return 10*d[Len] >= 9*Len;
}

int solve(char *s)
{	int l = 1, r = strlen(s+1);
	while (l <= r)
	{	int mid = (l+r) >> 1;
		if (check(s, mid)) l = mid + 1;
		else r = mid - 1;
	}
	return r;
}

char str[N];

int main()
{	scanf("%d %d", &n, &m);
	tzw.init();
	for (register int i = 1; i <= m; i++)
	{	scanf("%s", str);
		tzw.insert(str);
	}
	for (register int i = 1; i <= n; i++)
	{	scanf("%s", str+1);
		tzw.work(str);
		if(!check(str, 1)) puts("0");
		else printf("%d\n", solve(str));
	}
	return 0;
}

备注

注释里的坑我全踩了

posted @ 2018-04-04 17:01  EZ_WYC  阅读(202)  评论(0编辑  收藏  举报