題目鏈接

題意

$N$個串,每次詢問區間$[L,R]$中有多少子串$S_i$

思路

把$N$個串合成一個長字符串,對這個長字符串求后綴數組,包含$S_i$的子串的$height$連續排列,可以用線段樹維護height對應原串的位置,每次詢問變成找區間$[L^{\prime },R^{\prime }](lcp(L,R)>=S_i)$中有多少符合位置在$[L,R]$中

對$N$個串建$AC$自動機,$fail$樹上的每個父節點是當前節點的最長后綴,子串$S_i$的數量就是$S_i$的子樹的大小,在求出$fail$樹和$dfs$序之后,對于區間$[L,R]$的限制可以用主席樹或是離線處理+樹狀數組處理

#include <bits/stdc++.h> const int maxn = 4e5 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; char s[maxn], t[maxn]; int idx[maxn], start[maxn], len[maxn]; struct ac{int cntA[maxn], cntB[maxn], A[maxn], B[maxn];int Sa[maxn], tSa[maxn], height[maxn], Rank[maxn];int n, dp[maxn][21];void init(char *buf, int len) {n = len;for (int i = 0; i < 256; ++i) cntA[i] = 0;for (int i = 1; i <= n; ++i) cntA[(int)buf[i]]++;for (int i = 1; i < 256; ++i) cntA[i] += cntA[i-1];for (int i = n; i >= 1; --i) Sa[cntA[(int)buf[i]]--] = i;Rank[Sa[1]] = 1;for (int i = 2; i <= n; ++i) {Rank[Sa[i]] = Rank[Sa[i-1]];if (buf[Sa[i]] != buf[Sa[i-1]]) Rank[Sa[i]]++;}for (int l = 1; Rank[Sa[n]] < n; l <<= 1) {for (int i = 0; i <= n; ++i) cntA[i] = 0;for (int i = 0; i <= n; ++i) cntB[i] = 0;for (int i = 1; i <= n; ++i) {cntA[A[i] = Rank[i]]++;cntB[B[i] = (i+l<=n) ? Rank[i+l]:0]++;}for (int i = 1; i <= n; ++i) cntB[i] += cntB[i-1];for (int i = n; i >= 1; --i) tSa[cntB[B[i]]--] = i;for (int i = 1; i <= n; ++i) cntA[i] += cntA[i-1];for (int i = n; i >= 1; --i) Sa[cntA[A[tSa[i]]]--] = tSa[i];Rank[Sa[1]] = 1;for (int i = 2; i <= n; ++i) {Rank[Sa[i]] = Rank[Sa[i-1]];if (A[Sa[i]] != A[Sa[i-1]] || B[Sa[i]] != B[Sa[i-1]]) Rank[Sa[i]]++;}}for (int i = 1, j = 0; i <= n; ++i) {if (j) --j;int tmp = Sa[Rank[i] - 1];while (i+j <= n && tmp+j <= n && buf[i+j] == buf[tmp+j]) ++j;height[Rank[i]] = j;}} void st() {for (int i = 1; i <= n; ++i) dp[i][0] = height[i];for (int j = 1; j <= log2(n); ++j) {for (int i = 1; i + (1<<j)-1 <= n; ++i) {dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);}} }int rmq(int l, int r) {if (l == r) return inf;l++;int len = r - l + 1;int x = log2(len);return min(dp[l][x], dp[r-(1<<x)+1][x]);}vector<int> g[maxn<<2];void build(int rt, int l, int r) {for (int i = l; i <= r; ++i) g[rt].emplace_back(idx[Sa[i]]);sort(g[rt].begin(), g[rt].end());if (l == r) return;int mid = (l + r) >> 1;build(rt<<1, l, mid);build(rt<<1|1, mid+1, r);} int getL(int len, int t, int l, int r) {while (l <= r) {int mid = (l + r) >> 1;if (rmq(mid, t) >= len) r = mid - 1;else l = mid + 1;}return l;}int getR(int len, int t, int l, int r) {while (l <= r) {int mid = (l + r) >> 1;if (rmq(t, mid) >= len) l = mid + 1;else r = mid - 1;}return r;}int query(int rt, int l, int r, int ql, int qr, int x, int y) {if (l > qr || r < ql) return 0;if (l >= ql && r <= qr) {int L = lower_bound(g[rt].begin(), g[rt].end(), x) - g[rt].begin();int R = upper_bound(g[rt].begin(), g[rt].end(), y) - g[rt].begin();return R - L; }int mid = (l + r) >> 1;int sum = 0;if (mid >= ql) sum += query(rt<<1, l, mid, ql, qr, x, y);if (mid < qr) sum += query(rt<<1|1, mid+1, r, ql , qr, x, y);return sum;}void solve(int q) {build(1, 1, n);while (q--) {int l, r, k;scanf("%d %d %d", &l, &r, &k);int L = getL(len[k], Rank[start[k]], 1, Rank[start[k]]);int R = getR(len[k], Rank[start[k]], Rank[start[k]], n);printf("%d\n", query(1, 1, n, L, R, l, r));}} }S; int main() {int n, m, now = 0;scanf("%d %d", &n, &m);s[now] = '|';for (int i = 1; i <= n; ++i) {scanf("%s", t);len[i] = strlen(t);start[i] = now+1;for (int j = 0; j < len[i]; ++j) {s[++now] = t[j];idx[now] = i;}s[++now] = '|';idx[now] = 0;}s[now+1] = 0;S.init(s, now);S.st();S.solve(m);return 0; } #include <bits/stdc++.h> const int maxn = 5e5 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; vector<int> g[maxn]; struct Trie{int nex[maxn][26], fail[maxn], end[maxn];int pre[maxn];int L[maxn], R[maxn], cnt;int root, p;inline int newnode() {for (int i = 0; i < 26; ++i) nex[p][i] = -1;return p++;}inline void init() {cnt = p = 0;root = newnode();}void insert(int id, char *buf) {int now = root;for (int i = 0; buf[i]; ++i) {if (nex[now][buf[i]-'a'] == -1) nex[now][buf[i]-'a'] = newnode();pre[nex[now][buf[i]-'a']] = now;now = nex[now][buf[i]-'a'];}end[id] = now;}inline void build() {queue<int> que;fail[root] = root;for (int i = 0; i < 26; ++i) {if (nex[root][i] == -1) nex[root][i] = root;else {fail[nex[root][i]] = root;que.push(nex[root][i]);}}while (!que.empty()) {int now = que.front();que.pop();g[fail[now]].push_back(now);for (int i = 0; i < 26; ++i) {if (nex[now][i] == -1) nex[now][i] = nex[fail[now]][i];else {fail[nex[now][i]] = nex[fail[now]][i];que.push(nex[now][i]);}}}}void dfs(int x) {L[x] = ++cnt;for (auto it : g[x]) dfs(it);R[x] = ++cnt;} }S; struct ac{int l, r, id; }; int c[maxn]; void add(int x, int d) {while (x < maxn) {c[x]++;x += x&(-x);} } int getsum(int x) {int ans = 0;while(x) {ans += c[x];x -= x&(-x);}return ans; } int ans[maxn]; vector<ac> Q1[maxn], Q2[maxn]; char s[maxn]; int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n, m;scanf("%d %d", &n, &m);S.init();for (int i = 1; i <= n; ++i) {scanf("%s", s);S.insert(i, s);}S.build();S.dfs(0);for (int i = 1,l,r,k; i <= m; ++i) {scanf("%d %d %d", &l, &r, &k);Q1[l].push_back({S.L[S.end[k]], S.R[S.end[k]], i});Q2[r].push_back({S.L[S.end[k]], S.R[S.end[k]], i});}for (int i = 1; i <= n; ++i) {int now = S.end[i];while (now) {add(S.L[now], 1);now = S.pre[now];}for (int j = 0; j < (int)Q1[i+1].size(); ++j) {int l = Q1[i+1][j].l;int r = Q1[i+1][j].r;int id = Q1[i+1][j].id;ans[id] -= getsum(r) - getsum(l-1);}}for (int i = 1; i <= n; ++i) {for (int j = 0; j < (int)Q1[i].size(); ++j) {int l = Q1[i][j].l;int r = Q1[i][j].r;int id = Q1[i][j].id;ans[id] += getsum(r) - getsum(l-1);}}fill(c, c+maxn, 0);for (int i = n; i >= 1; --i) {int now = S.end[i];while (now) {add(S.L[now], 1);now = S.pre[now];}for (int j = 0; j < (int)Q2[i-1].size(); ++j) {int l = Q2[i-1][j].l;int r = Q2[i-1][j].r;int id = Q2[i-1][j].id;ans[id] -= getsum(r) - getsum(l-1);}}for (int i = 1; i <= m; ++i) {printf("%d\n", ans[i]);}return 0; }

總結

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