HDU 4416 Good Article Good sentence 后缀自动机

求字符串A与B，求不为B子串的A子串的个数。

与SPOJ 1812反过来。。
记得数据要沿Parent树更新。。

#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define FOR(i,j,k) for(i=j;i<=k;++i)const int rt = 1, N = 200005;int last = 1, cnt = 1, len = 0;int ch[N][26], fa[N], ma[N], rr[N], b[N], bucket[N >> 1];char str[N >> 1];void add(char c) {    int np = ++cnt, p = last, q, nq; last = np; ma[np] = ++len; rr[np] = 0;    memset(ch[np], 0, sizeof ch[np]);    while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];    if (!p) fa[np] = rt;    else {        q = ch[p][c];        if (ma[q] == ma[p] + 1) fa[np] = q;        else {            nq = ++cnt; memcpy(ch[nq], ch[q], sizeof ch[q]);            ma[nq] = ma[p] + 1; rr[nq] = 0;            fa[nq] = fa[q]; fa[q] = fa[np] = nq;            while (p && ch[p][c] == q) ch[p][c] = nq, p = fa[p];        }    }}int main() {    int i, n, t, p, y, c, Q, kase = 0; long long ans;    scanf("%d", &Q);    while (Q--) {        scanf("%d", &t); last = rt; cnt = 1; ans = len = 0;        memset(ch[rt], 0, sizeof ch[rt]);        scanf("%s", str); n = strlen(str);        for (i = 0; str[i]; ++i) add(str[i] - 'a');        FOR(i,0,n) bucket[i] = 0;        FOR(i,1,cnt) ++bucket[ma[i]];        FOR(i,1,n) bucket[i] += bucket[i - 1];        for (i = cnt; i; --i) b[bucket[ma[i]]--] = i;        while (t--) {            scanf("%s", str); p = rt; y = 0;            for (i = 0; str[i]; ++i) {                c = str[i] - 'a';                if (ch[p][c]) {                    p = ch[p][c];                    rr[p] = max(rr[p], ++y);                } else {                    while (p && !ch[p][c]) p = fa[p];                    if (p) y = ma[p] + 1, p = ch[p][c], rr[p] = max(rr[p], y);                    else p = rt, y = 0;                }            }        }        for(i=cnt;i;--i) if (rr[b[i]]) {            rr[fa[b[i]]] = max(rr[fa[b[i]]], rr[b[i]]);            if (rr[b[i]] < ma[b[i]]) ans += ma[b[i]] - rr[b[i]];        } else ans += ma[b[i]] - ma[fa[b[i]]];        printf("Case %d: %lld\n", ++kase, ans);    }    return 0;}

Good Article Good sentence

Problem Description

In middle school, teachers used to encourage us to pick up pretty sentences so that we could apply those sentences in our own articles. One of my classmates ZengXiao Xian, wanted to get sentences which are different from that of others, because he thought the distinct pretty sentences might benefit him a lot to get a high score in his article.
Assume that all of the sentences came from some articles. ZengXiao Xian intended to pick from Article A. The number of his classmates is n. The i-th classmate picked from Article Bi. Now ZengXiao Xian wants to know how many different sentences she could pick from Article A which don’t belong to either of her classmates?Article. To simplify the problem, ZengXiao Xian wants to know how many different strings, which is the substring of string A, but is not substring of either of string Bi. Of course, you will help him, won’t you?

Input

The first line contains an integer T, the number of test data.
For each test data
The first line contains an integer meaning the number of classmates.
The second line is the string A;The next n lines,the ith line input string Bi.
The length of the string A does not exceed 100,000 characters , The sum of total length of all strings Bi does not exceed 100,000, and assume all string consist only lowercase characters ‘a’ to ‘z’.

Output

For each case, print the case number and the number of substrings that ZengXiao Xian can find.

Sample Input

3
2
abab
ab
ba
1
aaa
bbb
2
aaaa
aa
aaa

Sample Output

Case 1: 3
Case 2: 3
Case 3: 1

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

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