# 【BZOJ3942】[Usaco2015 Feb]Censoring

## Description

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.

## Input

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).

whatthemomooofun
moo

## Sample Output

whatthefun

#include<stdio.h>
#include<string.h>
unsigned long long hs[1000010],ht;
unsigned long long seed[1000010];
int n,m;
int top;
char S[1000010],stack[1000010],T[1000010];
void BKDR()
{
seed[1]=131;
ht=T[0];
int i;
for(i=1;i<m;i++)
{
ht=ht*seed[1]+T[i];
seed[i+1]=seed[i]*seed[1];
}
}
int main()
{
scanf("%s%s",S,T);
n=strlen(S);
m=strlen(T);
BKDR();
int i,j;
for(i=0;i<m;i++)
{
stack[top++]=S[i];
hs[top]=hs[top-1]*seed[1]+S[i];
}
for(i=m;i<n;i++)
{
while(top>=m&&hs[top]-hs[top-m]*seed[m]==ht)
for(j=0;j<m;j++)
stack[--top]='\0';
stack[top++]=S[i];
hs[top]=hs[top-1]*seed[1]+S[i];
}
while(top>=m&&hs[top]-hs[top-m]*seed[m]==ht)
for(j=0;j<m;j++)
stack[--top]='\0';
printf("%s",stack);
return 0;
}

# 【BZOJ3940】[Usaco2015 Feb]Censoring

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int maxn=100010;
struct node
{
int fail,ch[26],cnt;
}p[maxn];
char str[maxn],w[maxn],ans[maxn];
int n,tot,len,sum;
int pos[maxn],q[maxn],l,r;
void build()
{
int i,u,t;
l=1;
q[++r]=1;
while(l<=r)
{
u=q[l++];
for(i=0;i<26;i++)
{
if(!p[u].ch[i])
{
if(u==1)    p[u].ch[i]=1;
else    p[u].ch[i]=p[p[u].fail].ch[i];
continue;
}
q[++r]=p[u].ch[i];
if(u==1)
{
p[p[u].ch[i]].fail=1;
continue;
}
t=p[u].fail;
while(!p[t].ch[i]&&t)    t=p[t].fail;
if(t)    p[p[u].ch[i]].fail=p[t].ch[i];
else    p[p[u].ch[i]].fail=1;
}
}
}
void search()
{
int i,j,u,t;
u=1;
pos[0]=1;
for(i=0;i<len;i++)
{
ans[++sum]=str[i];
pos[sum]=p[pos[sum-1]].ch[str[i]-'a'];
if(p[pos[sum]].cnt)    sum-=p[pos[sum]].cnt;
}
for(i=1;i<=sum;i++)    putchar(ans[i]);
}
int main()
{
scanf("%s",str);
scanf("%d",&n);
len=strlen(str);
int i,j,k,u;
tot=1;
for(i=1;i<=n;i++)
{
scanf("%s",w);
k=strlen(w);
u=1;
for(j=0;j<k;j++)
{
if(!p[u].ch[w[j]-'a'])    p[u].ch[w[j]-'a']=++tot;
u=p[u].ch[w[j]-'a'];
}
p[u].cnt=k;
}
build();
search();
}
posted @ 2017-01-06 16:17 CQzhangyu 阅读(...) 评论(...) 编辑 收藏