SCU4438-Censort

题目链接：http://acm.scu.edu.cn/soj/problem.action?id=4438

题意：给出两个字符串A, B,  要求每次将B中第一次出现的A删除, 然后将B剩下的两段前后合并, 然后重复这个过程直到没有可以删除的A位置, 问最后剩下的串是什么

解题思路：链表+kmp，对字符串A做kmp，然后和字符串B进行匹配

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f3f;

char a[5000010];
char b[5000010];
int nt[5000010];
int nex[5000010];
int p[5000010];
int len1, len2;

void getNext()
{
nt[0] = -1;
for (int i = 0; i < len1; i++)
{
int k=nt[i];
while (k>0&&	a[i]!=a[k])
k = nt[k];
nt[i + 1] = k + 1;
}
}

int main()
{
while (~scanf("%s%s",a,b))
{
len1 = strlen(a), len2 = strlen(b);
if (len1 > len2)
{
printf("%s\n", b);
continue;
}
getNext();
for (int i = 0; i <= len2; i++)
{
nex[i] = i + 1;
p[i] = i - 1;
}
for (int i = 0, j = 0; i < len2;)
{
if (j < len1 && b[i] == a[j]) j++;
else
{
while (j >=0&&b[i]!=a[j])
j = nt[j];
j++;
}
if (j == len1)
{
int x = i;
for (int k = 0; k < len1; k++) i = p[i];
if (i == -1)
{
p[nex[x]] = -1;
i = nex[x];
j = 0;
continue;
}
else
{
nex[i] = nex[x];
p[nex[x]] = i;
for (int k = 1;k<len1 ; k++)
{
if (p[i] != -1) i = p[i];
else break;
}
j = 0;
}
}
else i = nex[i];
}
int cnt=0,k = len2;
while (p[k] != -1)
{
a[cnt++] = b[p[k]];
k = p[k];
}
for (int i = cnt - 1; i >= 0; i--) printf("%c", a[i]);
printf("\n");
}
return 0;
}