原题链接:信息学奥赛一本通T1667-巧克力棒
解题思路:
AC自动机。使用栈保存匹配过程中的信息,若匹配成功弹出模式串,最后栈剩下的就是匹配完剩下的字符。
参考代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 | constexpr auto Inf = 0X3F3F3F3F ; #ifndef LOCAL #include <bits/stdc++.h> #endif typedef unsigned long long LL; using namespace std; namespace IO { inline LL read() { LL o = 0 , f = 1 ; char c = getchar(); while (c < '0' || c > '9' ) { if (c = = '-' ) f = - 1 ; c = getchar(); } while (c > '/' && c < ':' ) { o = o * 10 + c - '0' ; c = getchar(); } return o * f; } inline char recd() { char o; while ((o = getchar()) < '0' || o > '9' ); return o; } inline double reod() { double o = read(), f = 1 ; char c; while ((c = getchar()) > '/' && c < ':' ) o + = (c - '0' ) / (f * = 10 ); return o; } } using namespace IO; const int SIZE = 1E5 + 7 , Mod = ( 1E9 + 7 , 998244353 ); struct Aho { struct Node { int sub[ 26 ]; int fail, g; } Node[SIZE << 2 ]; int Ind; void clear() { for ( int pos = 0 ; pos < = Ind; pos + + ) { for ( int e = 0 ; e < 26 ; e + + ) Node[pos].sub[e] = 0 ; Node[pos].fail = Node[pos].g = 0 ; } Ind = 0 ; } int Ins(char S[], int g) { int now = 0 ; for ( int pos = 1 ; S[pos]; pos + + ) { if (!Node[now].sub[S[pos] - 'a' ]) Node[now].sub[S[pos] - 'a' ] = + + Ind; now = Node[now].sub[S[pos] - 'a' ]; } Node[now].g = g; return now; } void BUILD() { queue< int > que; for ( int pos = 0 ; pos < 26 ; pos + + ) if (Node[ 0 ].sub[pos]) Node[Node[ 0 ].sub[pos]].fail = 0 , que.push(Node[ 0 ].sub[pos]); while (!que.empty()) { int now = que.front(); que.pop(); for ( int pos = 0 ; pos < 26 ; pos + + ) { if (Node[now].sub[pos]) Node[Node[now].sub[pos]].fail = Node[Node[now].fail].sub[pos], que.push(Node[now].sub[pos]); else Node[now].sub[pos] = Node[Node[now].fail].sub[pos]; } } } } Aho; char T[SIZE], S[SIZE]; int stacks[SIZE], top, back[SIZE]; int main() { scanf( "%s" , T + 1 ); int N = read(); for ( int pos = 1 ; pos < = N; pos + + ) scanf( "%s" , S + 1 ), Aho.Ins(S, strlen(S + 1 )); Aho.BUILD(); int now = 0 ; for ( int pos = 1 ; T[pos]; pos + + ) { now = Aho.Node[now].sub[T[pos] - 'a' ]; back[pos] = now, stacks[ + + top] = pos; if (Aho.Node[now].g) { top - = Aho.Node[now].g; now = back[stacks[top]]; } } for ( int pos = 1 ; pos < = top; pos + + ) printf( "%c" , T[stacks[pos]]); } |
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