Codeforces 235C Cyclical Quest 后缀自动机
题目大意：
就是现在对于一个长度不超过10^6的字符串S, 接下来有n次询问(n <= 10^6), 每次询问有一个字符串x, 求x或者x轮换变换之后能在S中找到的对应字串个数, 输入的x的字符串总长度不超过10^6
轮换变化x比如 x = "aaba", 那么对应的要找出"aaba", "abaa", "baaa", "aaab"在S中出现次数的和
大致思路：
后缀自动机一发AC...
首先对于S建立后缀自动机, 然后对于每一个询问的字符串xi, 变成xi+xi的形式, 这样对于xi + xi这个字符串在S的后缀自动机上进行匹配, 当匹配长度到xi的长度时就将匹配长度减少一位, 如果当前匹配长度cnt在状态t上且Min(t) == cnt, 就转移到parent点, cnt--, 如果不是就cnt--, 依旧保持在这个t状态的位置,没到达一个cnt = length(xi)的状态, 这个状态的right集合的大小就是这样一个轮换出现的次数, 这样每次O(length(xi))匹配之后就得到了所有轮换的串出现的次数, 在匹配的时候记录已经计算了的状态不要重复计算就可以了
代码如下：
Result : Accepted Memory : 265180 KB Time : 280 ms
/*
* Author: Gatevin
* Created Time: 2015/4/11 15:21:05
* File Name: Rin_Tohsaka.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)
#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl
#define maxm 1000010
#define maxn 2000010
struct Suffix_Automation
{
struct State
{
State *par;
State *go[26];
int val, right, mi, cnt, vis;
void init(int _val = 0)
{
par = 0, val = _val, right = mi = cnt = vis = 0;
memset(go, 0, sizeof(go));
}
int calc()
{
if(par == 0) return 0;
else return val - par->val;
}
};
State *root, *last, *cur;
State nodePool[maxn];
State* newState(int val = 0)
{
cur->init(val);
return cur++;
}
void initSAM()
{
cur = nodePool;
root = newState();
last = root;
}
void extend(int w)
{
State *p = last;
State *np = newState(p->val + 1);
np->right = 1;
while(p && p->go[w] == 0)
{
p->go[w] = np;
p = p->par;
}
if(p == 0)
{
np->par = root;
}
else
{
State *q = p->go[w];
if(q->val == p->val + 1)
{
np->par = q;
}
else
{
State *nq = newState(p->val + 1);
memcpy(nq->go, q->go, sizeof(q->go));//刚开始这句写掉了...调试了一段时间才过样例...
nq->par = q->par;
q->par = nq;
np->par = nq;
while(p && p->go[w] == q)
{
p->go[w] = nq;
p = p->par;
}
}
}
last = np;
}
int d[maxm];
State* b[maxn];
void topo()
{
int maxVal = 0;
memset(d, 0, sizeof(d));
int cnt = cur - nodePool;
for(int i = 1; i < cnt; i++)
maxVal = max(maxVal, nodePool[i].val), d[nodePool[i].val]++;
for(int i = 1; i <= maxVal; i++) d[i] += d[i - 1];
for(int i = 1; i < cnt; i++) b[d[nodePool[i].val]--] = &nodePool[i];
b[0] = root;
}
void SAMInfo()
{
State *p;
int cnt = cur - nodePool;
for(int i = cnt - 1; i > 0; i--)
{
p = b[i];
p->par->right += p->right;
p->mi = p->par->val + 1;
}
}
void solve(char *x, int len, int L, int cas)
{
State *now = root;
int cnt = 0;
int ans = 0;
for(int i = 0; i < L - 1; i++)
{
int w = x[i] - 'a';
if(now->go[w])
{
now = now->go[w];
cnt++;
if(cnt == len)
{
if(now->vis != cas)
{
ans += now->right;
now->vis = cas;//标记这个状态计算过了
}
if(len == now->mi)//转移到其Parent点
{
now = now->par;
cnt = now->val;
}
else
cnt--;//保留在这个点
}
}
else
{
while(now && now->go[w] == 0)
{
now = now->par;
if(now)
cnt = now->val;
}
if(now == 0)
{
now = root;
cnt = 0;
}
else
{
now = now->go[w];
cnt++;
if(cnt == len)
{
if(now->vis != cas)
{
ans += now->right;
now->vis = cas;//标记计算过了
}
if(len == now->mi)
{
now = now->par;
cnt = now->val;
}
else
cnt--;
}
}
}
}
printf("%d\n", ans);
}
};
Suffix_Automation sam;
char s[maxm];
int n;
char x[maxn];
int main()
{
gets(s);
int lens = strlen(s);
sam.initSAM();
for(int i = 0; i < lens; i++)
sam.extend(s[i] - 'a');
sam.topo();
sam.SAMInfo();
scanf("%d", &n);
getchar();
for(int cas = 1; cas <= n; cas++)
{
gets(x);
int len = strlen(x);
for(int i = 0; i < len; i++)
x[i + len] = x[i];
sam.solve(x, len, 2*len, cas);
}
return 0;
}