Censoring
传送门：链接 来源：UPC8203
题目描述
Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inappropriate article on how to cook the perfect steak, which FJ would rather his cows not see (clearly, the magazine is in need of better editorial oversight).
FJ has taken all of the text from the magazine to create the string S of length at most 10^6 characters. From this, he would like to remove occurrences of a substring T to censor the inappropriate content. To do this, Farmer John finds the _first_ occurrence of T in S and deletes it. He then repeats the process again, deleting the first occurrence of T again, continuing until there are no more occurrences of T in S. Note that the deletion of one occurrence might create a new occurrence of T that didn't exist before.
Please help FJ determine the final contents of S after censoring is complete.
输入
The first line will contain S. The second line will contain T. The length of T will be at most that of S, and all characters of S and T will be lower-case alphabet characters (in the range a..z).
输出
The string S after all deletions are complete. It is guaranteed that S will not become empty during the deletion process.
样例输入
whatthemomooofun
moo
样例输出
whatthefun
题目大意：
给出a，p两个字符串从a中消去所有的p，直到不能消去为止。
思路分析：
尝试了三种做法：
1、直接vector删除，结果TLE。
2、Next[]数组+vector删除，结果TLE（应该是哪里bug死循环了，现在也没找到原因，估计是kmp的代码不行）。
3、看了别人的题解：传送门，发现可以用堆栈模拟来做。（下面讲的就是这种方法）
之前一想到删除操作第一反应就是vector，看来以后也要考虑堆栈了，用vector应该也行。
看上面博客写的kmp代码和我的不太一样，顺便学习了一下他的代码。
AC代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MAX=1e6;
LL Next[MAX+5],flag[MAX+5];
char Stack[MAX+5];
void getNext(string p,LL lp,LL Next[]){
LL k=0;
for(LL i=1;i<lp;i++){
while(k&&p[i]!=p[k]) k=Next[k];
if(p[i]==p[k]) k++;
Next[i+1]=k;
}
}
void kmp(string a,string p,LL la,LL lp){
LL j=0,top=0;
for(LL i=0;i<la;i++){
Stack[++top]=a[i];
while(j&&a[i]!=p[j]) j=Next[j];
if(a[i]==p[j]) j++;
if(j==lp) top-=lp,j=flag[top];
flag[top]=j;
}
for(int i=1;i<=top;i++){ ///堆栈中剩余的都是没被删除的，直接输出就行了。
cout<<Stack[i];
}
cout<<endl;
}
int main()
{
string a,p;
cin>>a>>p;
LL la=a.size(),lp=p.size();
getNext(p,lp,Next);
kmp(a,p,la,lp);
return 0;
}