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;
}

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