Codeforces 235C

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題目：給定一主串\(S\)，\(n\)次詢問，每次詢問串\(t\)的所有循環移位串的出現的次數和

做法：建\(SAM\)，對于詢問串\(t\)，將他復制一份放在后邊，在后綴自動機上匹配，如果匹配長度大于\(|t|\)，就沿著\(fa\), 找到第一次大于\(|t|\)的位置，用這個狀態的\(right\)數組更新答案。注意到可能會匹配到重復的狀態，所以要對會更新答案的狀態去重。

#include <bits/stdc++.h> typedef long long ll; const int N = 1000010 << 1; using namespace std;struct SAM{int n, step[N], fa[N], ch[N][26], right[N], num[N], last, root, cnt;char s[N>>1];SAM(){last = root = ++cnt; }void add(int x){int tmp = s[x] - 'a', p = last, np = ++cnt;step[last = np] = x, right[np] = 1;while(p && !ch[p][tmp]) ch[p][tmp] = np, p = fa[p];if(!p) fa[np] = root;else{int q = ch[p][tmp];if(step[q] == step[p] + 1) fa[np] = q;else{int nq = ++cnt; step[nq] = step[p] + 1;memcpy(ch[nq], ch[q], sizeof(ch[q]));fa[nq] = fa[q], fa[q] = fa[np] = nq;while(ch[p][tmp] == q) ch[p][tmp] = nq, p = fa[p];}}}int tmp[N];void calright(){memset(tmp, 0, sizeof(tmp));for(int i = 1; i <= cnt; i++) tmp[step[i]]++;for(int i = 1; i <= n; i++) tmp[i] += tmp[i - 1];for(int i = cnt; i; i--) num[tmp[step[i]]--] = i;for(int i = cnt; i; i--) right[fa[num[i]]] += right[num[i]];}void run(){scanf(" %s",s+1), n = strlen(s + 1);for(int i = 1; i <= n; i++) add(i);calright();}char str[N];int len;set<int> S;ll solve() {scanf(" %s",str+1), len = strlen(str+1);for(int i = 1; i <= len; ++i) str[i+len] = str[i];int now = root, tmp = 0; ll ans = 0;for(int i = 1; i <= len+len; ++i) {int x = str[i]-'a';if(ch[now][x]) now = ch[now][x], ++tmp;else {while(now && !ch[now][x]) now = fa[now];if(!now) now = root, tmp = 0;else tmp = step[now]+1, now = ch[now][x];}while(now!=1 && step[fa[now]] >= len) now = fa[now], tmp = step[now];if(tmp >= len) S.insert(now);}for(set<int>::iterator it = S.begin(); it != S.end(); ++it) ans += right[*it];S.clear();return ans;} } Fe;int main() {Fe.run();int q;scanf("%d",&q);while(q--) {printf("%I64d\n",Fe.solve());} }

轉載于:https://www.cnblogs.com/RRRR-wys/p/10299830.html

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