loj#6031. 「雅礼集训 2017 Day1」字符串(SAM 广义SAM 数据分治)
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2022-04-28 13:57:15
题意 "链接" Sol $10^5$次询问每次询问$10^5$个区间。。这种题第一感觉就是根号/数据分治的模型。 $K$是个定制这个很关键。 考虑$K$比较小的情况,可以直接暴力建SAM,$n^2$枚举$w$的子串算出现次数。询问用个$n^2$的vector记录一下每次在vector里二分就好。 $ ......
题意
sol
\(10^5\)次询问每次询问\(10^5\)个区间。。这种题第一感觉就是根号/数据分治的模型。
\(k\)是个定制这个很关键。
考虑\(k\)比较小的情况,可以直接暴力建sam,\(n^2\)枚举\(w\)的子串算出现次数。询问用个\(n^2\)的vector记录一下每次在vector里二分就好。
\(k\)比较大的情况我没想到什么好的做法,网上的做法复杂度也不是很好。。
然后写了个广义sam + 暴力跳parent就过了。。
不过这题思想还是很好的
#include<bits/stdc++.h> #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define ll long long #define ull unsigned long long #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 4e5 + 10, inf = 1e9 + 1, mod = 1e9 + 7; const double eps = 1e-9, pi = acos(-1); template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int n, m, q, k; char s[maxn], w[maxn]; string q[maxn]; int l[maxn], r[maxn], a[maxn], b[maxn], pos[maxn]; vector<int> line[1001][1001], v[maxn]; int ch[maxn][26], siz[maxn], len[maxn], fa[maxn], las = 1, root = 1, tot = 1; void insert(int x, int opt) { int now = ++tot, pre = las; las = now; len[now] = len[pre] + 1; siz[now] = opt; for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now; if(!pre) fa[now] = root; else { int q = ch[pre][x]; if(len[pre] + 1 == len[q]) fa[now] = q; else { int nq = ++tot; len[nq] = len[pre] + 1; fa[nq] = fa[q]; memcpy(ch[nq], ch[q], sizeof(ch[q])); fa[now] = fa[q] = nq; for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq; } } } void dfs(int x) { for(auto &to : v[x]) { dfs(to); siz[x] += siz[to]; } } int query(vector<int> &q, int a, int b) { return (upper_bound(q.begin(), q.end(), b) - lower_bound(q.begin(), q.end(), a)); } ll solve1(int a, int b) { ll ret = 0; for(int i = 1; i <= k; i++) { int now = root; for(int j = i; j <= k; j++) { int x = w[j] - 'a'; now = ch[now][x]; if(!now) break; else ret += 1ll * siz[now] * query(line[i][j], a, b); } } return ret; } void build() { for(int i = 1; i <= tot; i++) v[fa[i]].push_back(i); dfs(root); } signed main() { // freopen("string9.in", "r", stdin); freopen("b.out", "w", stdout); n = read(); m = read(); q = read(); k = read(); scanf("%s", s + 1); for(int i = 1; i <= n; i++) insert(s[i] - 'a', 1); if(k <= 1000) { build(); for(int i = 1; i <= m; i++) l[i] = read() + 1, r[i] = read() + 1, line[l[i]][r[i]].push_back(i); for(int i = 1; i <= q; i++) { scanf("%s", w + 1); int a = read() + 1, b = read() + 1; cout << solve1(a, b) << '\n'; } } else { for(int i = 1; i <= m; i++) l[i] = read() + 1, r[i] = read() + 1; for(int i = 1; i <= q; i++) { cin >> q[i]; a[i] = read() + 1, b[i] = read() + 1; las = 1; for(auto &x : q[i]) insert(x - 'a', 0); } build(); for(int i = 1; i <= q; i++) { memset(pos, 0, sizeof(pos)); int now = root; for(int j = 0; j < q[i].length(); j++) { int x = q[i][j] - 'a'; now = ch[now][x]; if(!now) break; else pos[j + 1] = now; } ll ans = 0; for(int j = a[i]; j <= b[i]; j++) { int cur = pos[r[j]]; while(len[fa[cur]] >= r[j] - l[j] + 1) cur = fa[cur];//这里可以卡掉 ans += siz[cur]; } cout << ans << '\n'; } } return 0; }