标签:void 字符串 -- second 后缀数组 inf tin title string
题目大意:求一个字符串中不同子串的个数。
裸的后缀数组
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; const int N = 1e5 + 7; const int M = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; char s[N]; int sa[N], t[N], t2[N], c[N], rk[N], height[N], id[N], b[N], d[N], n, tot; void buildSa(char *s, int n, int m) { int i, j = 0, k = 0, *x = t, *y = t2; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[i] = s[i]]++; for(i = 1; i < m; i++) c[i] += c[i - 1]; for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1) { int p = 0; for(i = n - k; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[y[i]]]++; for(i = 1; i < m; i++) c[i] += c[i - 1]; for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(int i = 1; i < n; i++) { if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) x[sa[i]] = p - 1; else x[sa[i]] = p++; } if(p >= n) break; m = p; } for(i = 1; i < n; i++) rk[sa[i]] = i; for(i = 0; i < n - 1; i++) { if(k) k--; j = sa[rk[i] - 1]; while(s[i + k] == s[j + k]) k++; height[rk[i]] = k; } } int main() { int T; scanf("%d", &T); while(scanf("%s", s) != EOF) { n = strlen(s); buildSa(s, n + 1, 180); LL ans = 0; for(int i = 1; i <= n; i++) { ans += (n - sa[i]) - height[i]; } printf("%lld\n", ans); } return 0; } /* */
SPOJ - SUBST1 D - New Distinct Substrings
标签:void 字符串 -- second 后缀数组 inf tin title string
原文地址:https://www.cnblogs.com/CJLHY/p/9343689.html