SCU 4438 Censor (KMP)

题目大意：

就是现在给出两个字符串A, B, 长度都不超过500万, 要求每次将B中第一次A的出现删除, 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

大致思路：

首先看这个字符串的长度不难想到用KMP处理出A的next数组

然后考虑在B上进行匹配, 这里我用的是两个数组来模拟链表, 表示剩下的串, 每次发生删除时, 将当前位置的指针前移即可, 由于删除的串不会在之后进行匹配, 整体的时间复杂度是O(|A| + |B|)

代码如下：

Result : Accepted Memory : 51820 KB Time : 848 ms

/** Author: Gatevin* Created Time:  2015/10/1 9:22:45* File Name: A.cpp*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
char a[5000010];
char b[5000010];
int next[5000010];
int len1, len2;
void getNext()
{for(int i = 0; i <= len1; i++)next[i] = 0;for(int i = 1; i < len1; i++){int j = i;while(j > 0){j = next[j];if(a[j] == a[i]){next[i + 1] = j + 1;break;}}}return;
}
int nex[5000010];
int bef[5000010];
int main()
{while(gets(a)){
<span style="white-space:pre"> </span>gets(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;bef[i] = i - 1;}for(int i = 0, j = 0; i < len2;){if(j < len1 && b[i] == a[j]) j++;else{while(j > 0){j = next[j];if(b[i] == a[j]){j++;break;}}}if(j == len1){int ti = i;for(int k = 0; k < len1; k++){i = bef[i];}if(i == -1){bef[nex[ti]] = -1;i = nex[ti];j = 0;continue;}else{nex[i] = nex[ti];bef[nex[ti]] = i;for(int k = 2; k < len1; k++){if(bef[i] != -1)i = bef[i];}j = 0;}}else i = nex[i];}int cnt = 0;int now = len2;while(bef[now] != -1){//putchar(b[bef[now]]);a[cnt++] = b[bef[now]];now = bef[now];}for(int i = cnt - 1; i >= 0; i--)putchar(a[i]);putchar('\n');}return 0;
}

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