2016 多校 Multi-University Training Contest 4 Substring
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hdu 5769 Substring
题意很简单,给你一个串和一个字符,问你含有这个字符的这个串的子串有几个
后缀自动机应该能使,但是不会,多校结束之后用后缀数组过掉的
一个串的连续子串的总数: sigma(len - height[I] - suffixarray[I])
那么带有该字符的只要填一点东西进去就可以求了
也就是在 (height[I] + suffixarray[I])和 该后缀后方第一个该字符的位置 这两个值里面选一个最大值就行了
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int SIZE = 100005;struct suffixArray{ int wa[SIZE], wb[SIZE], wv[SIZE], ws[SIZE]; int sa[SIZE];//后缀数组 int height[SIZE]; int rank[SIZE]; int r[SIZE];//转化后的数组,长度比源字符串长1 int len; int Max;//用于基数排序,是字符串中的最大值 //int n, m; void init() { memset(r, 0, sizeof r); } void clear() { Max = len = 0; memset(sa, 0, sizeof sa); memset(height, 0, sizeof height); memset(rank, 0, sizeof rank); memset(r, 0, sizeof r); } int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } //尾部自动添 0,并增加字符串长度(+1),为了基数排序不出错 int go(char *str) { int tmp = 0; Max = 0; for (int i = 0; str[i]; i++) { r[i] = str[i] - 'a' + 1; Max = max(Max, r[i]); tmp++; } r[tmp++] = 0; len = tmp; return tmp; } void doubleAdd() { //go(str); //Max = 27; int i, j, p; int n = len, m = Max + 1; int *x = wa, *y = wb, *t; // 对长度为1的串进行排序 for (i = 0; i < m; i++) ws[i] = 0; for (i = 0; i < n; i++) ws[x[i] = r[i]]++; for (i = 1; i < m; i++) ws[i] += ws[i - 1]; for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i; //j为当前字符串长度,y是第二关键字排序的结果 for (j = 1, p = 1; p < n; j *= 2, m = p) { for (p = 0, i = n - j; i < n; i++) y[p++] = i; for (i = 0; i < n; i++) { if (sa[i] >= j) y[p++] = sa[i] - j; } //对第一关键字进行排序 for (i = 0; i < n; i++) wv[i] = x[y[i]]; for (i = 0; i < m; i++) ws[i] = 0; for (i = 0; i < n; i++) ws[wv[i]]++; for (i = 1; i < m; i++) ws[i] += ws[i - 1]; for (i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i]; //到此求出新的sa值 //计算新的rank值,这里之前y已经没用了,就利用y来保存rank for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } return; } void calHeight() { int i, j, k = 0; for (i = 1; i < len; i++) rank[sa[i]] = i; for (i = 0; i < len; height[rank[i++]] = k) { //寻找当前长度的串排名前一位的串,这次求得height的值一定是上一次求得的height值-1 if (rank[i] > 0) for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++); else k = 0; } return; }}SA;int main(void){ //freopen("C:\\Users\\wLIEd\\Desktop\\in.txt", "r", stdin); char ch; char str[SIZE]; int T; int cas = 1; int len; long long ans; cin >> T; while (T--) { ans = 0; cin >> ch; cin >> str; SA.clear(); len = SA.go(str); SA.doubleAdd(); SA.calHeight(); for (int i = 1; i < len; i++) { for (int j = SA.sa[i]; j < len; j++) { if (str[j] == ch) { if (j <= SA.sa[i] + SA.height[i]) { ans += len - SA.sa[i] - SA.height[i] - 1; break; } else { ans += len - j - 1; break; } } } } cout << "Case #" << cas++ << ": " << ans << endl; } return 0;}
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