題目鏈接

題意

$Q$次詢問，每次詢問求$S$的子串出現(xiàn)$K$次的位置

思路

剛開始想的是AC自動機，但是建自動機會超時，后來學長想到后綴數(shù)組+主席樹的做法$Orz...$
出現(xiàn)$K$次的字符串，在后綴數(shù)組中連續(xù)出現(xiàn)，求出詢問對應(yīng)后綴前后滿足區(qū)間內(nèi)的$height>=len$的最長的區(qū)間，此時區(qū)間$[L,R]$對應(yīng)的位置都包含子串，主席樹找$[L,R]$中第K小的位置即可

#include <bits/stdc++.h> #define endl '\n' const int maxn = 1e5 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; char s[maxn]; struct Tree{int tot, rt[maxn], lson[maxn*20], rson[maxn*20], cnt[maxn*20];int build (int l, int r) {int o = ++tot;int mid = (l + r) >> 1;cnt[o] = 0;if (l != r) {lson[o] = build(l, mid);rson[o] = build(mid+1, r);}return o;}int update(int prev, int l, int r, int v) {int o = ++tot;int mid = (l + r) >> 1;lson[o] = lson[prev];rson[o] = rson[prev];cnt[o] = cnt[prev] + 1;if (l != r) {if (v <= mid) lson[o] = update(lson[o], l, mid, v);else rson[o] = update(rson[o], mid+1, r, v);}return o;}int query(int u, int v, int l, int r, int k) {if (l == r) return l;int mid = (l + r) >> 1;int num = cnt[lson[v]] - cnt[lson[u]];if (num >= k) return query(lson[u], lson[v], l, mid, k);else return query(rson[u], rson[v], mid+1, r, k-num);} }; struct SuffixArray{ // 下標1int cntA[maxn], cntB[maxn], A[maxn], B[maxn];int Sa[maxn], tsa[maxn], height[maxn], Rank[maxn]; // Sa[i] 排名第i的下標， Rank[i] 下標i的排名int n, dp[maxn][18];Tree T;void build() {T.tot = 0;T.rt[0] = T.build(1, n);for (int i = 1; i <= n; ++i) T.rt[i] = T.update(T.rt[i-1], 1, n, Sa[i]);}void init(char *buf, int len) { // 預(yù)處理，sa，rank，heightn = len;for (int i = 0; i < 128; ++i) cntA[i] = 0;for (int i = 1; i <= n; ++i) cntA[(int)buf[i]]++;for (int i = 1; i < 128; ++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]);}int findl(int pos, int len) {int l = 1, r = pos;while (l <= r) {int mid = (l + r) >> 1;if (rmq(mid, pos) < len) l = mid + 1;else r = mid - 1;}return l;}int findr(int pos, int len) {int l = pos, r = n;while (l <= r) {int mid = (l + r) >> 1;if (rmq(pos, mid) < len) r = mid - 1;else l = mid + 1;}return r;}int solve(int ql, int qr, int k) {int len = qr - ql + 1;int pos = Rank[ql];int l = findl(pos, len);int r = findr(pos, len);if (r - l + 1 < k) return -1;return T.query(T.rt[l-1], T.rt[r], 1, n, k);} }S; int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int T;scanf("%d", &T);while (T--) {int n, m;scanf("%d %d", &n, &m);scanf("%s", s+1);S.init(s, strlen(s+1));S.st();S.build();int l, r, k;for (int i = 0; i < m; ++i) {scanf("%d %d %d", &l, &r, &k);printf("%d\n", S.solve(l, r, k));} }return 0; } 與50位技術(shù)專家面對面20年技術(shù)見證，附贈技術(shù)全景圖

總結(jié)

以上是生活随笔為你收集整理的HDU - 6704 K-th occurrence (后缀数组+主席树)的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。