[USACO15FEB]Censoring S【KMP+單調棧】

題目鏈接

---

  我們要讓S串去刪除T串，且不斷的刪除存在過的T串，所以就是可以用一個單調棧維護一個存在的串，然後因爲可能存在刪除，所以我們要記錄每一位的最大匹配個數。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e6 + 7;
char S[maxN], T[maxN];
int len[2], nex[maxN], sam[maxN] = {0};
int Stap[maxN], Stop;
int main()
{
    scanf("%s%s", S, T);
    len[0] = (int)strlen(S); len[1] = (int)strlen(T);
    int k = -1; nex[0] = -1;
    for(int i=1; i<len[1]; i++)
    {
        while(~k && T[k + 1] ^ T[i])
        {
            k = nex[k];
        }
        if(T[k + 1] == T[i]) k++;
        nex[i] = k;
    }
    k = -1; Stop = 0;
    for(int i=0; i<len[0]; i++)
    {
        while(~k && S[i] ^ T[k + 1]) k = nex[k];
        if(T[k + 1] == S[i]) k++;
        sam[i] = k + 1;
        Stap[++Stop] = i;
        if(sam[i] == len[1])
        {
            Stop -= len[1];
            k = sam[Stap[Stop]] - 1;
        }
    }
    for(int i=1; i<=Stop; i++) printf("%c", S[Stap[i]]);
    printf("\n");
    return 0;
}