牛客多校第九場 Niuniu is practicing typing.（kmp優化）

鏈接：https://www.nowcoder.com/acm/contest/147/F
來源：牛客網
 

Niuniu is practicing typing.

Given n words, Niuniu want to input one of these. He wants to input (at the end) as few characters (without backspace) as possible,

to make at least one of the n words appears (as a suffix) in the text.

Given an operation sequence, Niuniu want to know the answer after every operation.

An operation might input a character or delete the last character.

輸入描述:

The first line contains one integer n.
In the following n lines, each line contains a word.
The last line contains the operation sequence.
'-' means backspace, and will delete the last character he typed.

He may backspace when there is no characters left, and nothing will happen.

1 <= n <= 4
The total length of n words <= 100000

The length of the operation sequence <= 100000

The words and the sequence only contains lower case letter.

輸出描述:

You should output L +1 integers, where L is the length of the operation sequence.

The i-th(index from 0) is the minimum characters to achieve the goal, after the first i operations.

示例1

輸入

複製

2
a
bab
baa-

輸出

複製

1
1
0
0
0

這個題就可以先考慮只有一個模式串，在一個模式串下我們和操作串進行匹配，如果在模式串的i位置失配，那麼跳到模式串的Next[i] 的位置，如果沒有刪除操作，那麼就是一個kmp的複雜度，可是現在可以刪除，理論上加一個棧記錄與模式串匹配的位置，可以實現回退。可是會有這樣一個問題。

模式串 ：aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

操作傳     aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab-c-v-d-s-d-a-w-........

這樣以來，失配以後每次都要大量的跳轉，非常費時間。因此考慮記錄每個位置失配以後，每個字母對應的應該跳到的位置，

這樣一來，每一次跳轉肯定都是0（1）啦。

因爲所給的要匹配的字符串長度隨時變化，可以用變量cur記錄字符串裏有幾個字符，pos[cur] 記錄第cur個字符在模式串的匹配位置。那麼如果是刪除操作，只需要cur -- ，字符減少一個。如果是加字符，只需要cur++，pos[cur] 紀錄新加字母后模式串匹配的位置。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5+666;

int n,Next[maxn],pre[maxn][26],Min[maxn],pos[maxn];
char s[5][maxn],sx[maxn];

void getNext(char s[])
{
	int len = strlen(s),i = 0;
	int k = -1;
	Next[0] = -1;
	while(i < len)
	{
		if(k == -1 || s[k] == s[i])
		{
			Next[++i] = ++k;
		}
		else
		{
			k = Next[k];
		}
	}
	for(int i = 0; i <= len; i++)
	{
	    for(int j = 0; j < 26; j++)
		    pre[i][j] = i == 0 ? 0 : pre[Next[i]][j];
		if(i < len) pre[i][s[i]-'a'] = i+1;
	}
}

void solve(char s[])
{
	getNext(s);
	int cur = 0;pos[0] = 0;
	int len = strlen(s);
	Min[0] = min(Min[0],len);
    for(int i = 1; sx[i]; i++)
    {
    	if(sx[i] == '-'){
    		cur = max(cur-1,0);
		}else
		{
			pos[cur] = pre[pos[cur++]][sx[i]-'a'];	
		}
		Min[i] = min(Min[i],len-pos[cur]);
	}
}

int main()
{
	scanf("%d",&n);
	for(int i = 1; i <= n; i++) scanf("%s",s[i]);
	scanf("%s",sx+1);
	int len = strlen(sx+1)+1;
	memset(Min,63,sizeof(Min));
	for(int i = 1; i <= n; i++) solve(s[i]);
	for(int i = 0; i < len; i++)
	{
		printf("%d\n",Min[i]);
	}
	return 0;
}