Bzoj 1212 [hnoi2004]l language "AC automata + backpack"

Topic link "http://www.lydsy.com/JudgeOnline/problem.php?id=1212"

Test instructions: Give you some words, and then give a text string with no punctuation, all lowercase characters. Now let's ask you to make a text string T with the given word, and ask for the longest common prefix of s and T.

Puzzle: AC automata + backpack, backpack dp[i], indicates whether the length of the "1,i" prefix, maintenance in the automaton Len[i], the node I to the root node distance, End[i], node I is the end of a word. In the query, we only need to jump on the corresponding trie, the time complexity of X * N*log (N).

#include <bits/stdc++.h>using namespace Std;const int maxn = 1024x768 * 1024x768 + 15;int dp[maxn];struct aho_c{int next[    MAXN][26], FAIL[MAXN], END[MAXN], LEN[MAXN];    int root, SZ;        int NewNode () {for (int i = 0; i <; i++) next[sz][i] = 1;        end[sz++] = 0;    return sz-1;        } void Init () {sz = 0;    root = NewNode ();        } void Insert (char buf[]) {int len = strlen (BUF);        int now = root; for (int i = 0; i < len; i++) {if (Next[now][buf[i]-' a '] = =-1) next[now][buf[i]-' A            '] = NewNode ();            now = next[now][buf[i]-' a '];        Len[now] = i + 1;    } end[now]++;        } void Build () {queue<int>q;        Fail[root] = root;            for (int i = 0; i < i++) if (next[root][i] = = 1) next[root][i] = root;                else {Fail[next[root][i]] = root; Q.push (NeXt[root][i]); } while (!            Q.empty ()) {int now = Q.front ();            Q.pop ();                 for (int i = 0; i < i++) if (next[now][i] = = 1) next[now][i] = Next[fail[now]][i];                    else {Fail[next[now][i]] = next[fail[now]][i];                Q.push (Next[now][i]);        }}} void Query (char buf[]) {int len = strlen (buf + 1);        int now = root;            for (int i = 1; I <= len; i++) {dp[i] = 0;            now = next[now][buf[i]-' a '];            int temp = now;            int tmp = len[temp]; while (temp! = root) {if (end[temp]) {int pos = i-len[temp]                    ;                Dp[i] = max (Len[temp] + Dp[pos], dp[i]);            } temp = fail[temp];  }}}} Ac;char Buf[maxn * 2];int main () {int N, M;  scanf ("%d%d", &n, &m);    Ac.init ();        for (int i = 1; I <= N; i++) {scanf ("%s", buf); Ac.    Insert (BUF); } AC.    Build ();        for (int i = 1; I <= M; i++) {scanf ("%s", buf + 1); Ac.        Query (BUF);        int len = strlen (buf + 1);        int ma = 0;                for (int i = len; I >= 1; i--) {if (dp[i] = = i) {ma = i;            Break    }} printf ("%d\n", MA); } return 0;}

　　

Bzoj 1212 [hnoi2004]l language "AC automata + backpack"