Searching the String
Time Limit: 7 Seconds Memory Limit: 129872 KB
Little jay really hates to deal with string. But moondy likes it very much, and she's so mischievous that she often gives jay some dull problems related to string. And one day, moondy gave jay another problem, poor jay finally broke out and cried, " Who can help me? I'll bg him! "
So what is the problem this time?
First, moondy gave jay a very long string A. Then she gave him a sequence of very short substrings, and asked him to find how many times each substring appeared in string A. What's more, she would denote whether or not founded appearances of this substring are allowed to overlap.
At first, jay just read string A from begin to end to search all appearances of each given substring. But he soon felt exhausted and couldn't go on any more, so he gave up and broke out this time.
I know you're a good guy and will help with jay even without bg, won't you?
Input
Input consists of multiple cases( <= 20 ) and terminates with end of file.
For each case, the first line contains string A ( length <= 10^5 ). The second line contains an integer N ( N <= 10^5 ), which denotes the number of queries. The next N lines, each with an integer type and a string a ( length <= 6 ), type = 0 denotes substring a is allowed to overlap and type = 1 denotes not. Note that all input characters are lowercase.
There is a blank line between two consecutive cases.
Output
For each case, output the case number first ( based on 1 , see Samples ).
Then for each query, output an integer in a single line denoting the maximum times you can find the substring under certain rules.
Output an empty line after each case.
Sample Input
ab
2
0 ab
1 ab
abababac
2
0 aba
1 aba
abcdefghijklmnopqrstuvwxyz
3
0 abc
1 def
1 jmn
Sample Output
Case 1
1
1
Case 2
3
2
Case 3
1
1
0
题意：给一个字符串s和n次询问，每次询问一个字符串在s中出现的次数，ord为0表示允许重叠，ord为1表示不允许重叠。
对于允许重叠的情况，建立好自动机之后s再上面跑一边就行了，不允许重叠需要记录该字串上一次出现的位置，没次与上次的位置比较判断是否合法。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
#define ll long long
const int mod = 100000;
const int maxm = 600006;
const int INF = 1e9 + 7;
int n;
int tr[maxm][26], fail[maxm], last[maxm], cnt;
int ord[maxm], id[maxm], pos[maxm], ans[maxm][2];
char str[maxm], ch[maxm];
void init()
{
memset(last, 0, sizeof(last));
memset(tr, 0, sizeof(tr));
memset(fail, 0, sizeof(fail));
memset(pos, -1, sizeof(pos));
memset(ans, 0, sizeof(ans));
cnt = 0;
}
int insert(char *s)
{
int len = strlen(s);
int now = 0;
for (int i = 0;i < len;i++)
{
int num = s[i] - 'a';
if (!tr[now][num])
tr[now][num] = ++cnt;
now = tr[now][num];
}
last[now] = len;
return now;
}
void find_fail()
{
int now;
queue<int>q;
for (int i = 0;i < 26;i++)
if (tr[0][i]) q.push(tr[0][i]);
while (!q.empty())
{
now = q.front();q.pop();
for (int i = 0;i < 26;i++)
{
if (tr[now][i])
{
fail[tr[now][i]] = tr[fail[now]][i];
q.push(tr[now][i]);
}
else tr[now][i] = tr[fail[now]][i];
}
}
}
void find(char *s)
{
int p = 0, len = strlen(s);
for (int i = 0;i < len;i++)
{
int j = s[i] - 'a';
p = tr[p][j];
if (p == 0) continue;
int q = p;
while (q)
{
if (last[q])
{
ans[q][0]++;
if (i - pos[q] >= last[q])
pos[q] = i, ans[q][1]++;
}
q = fail[q];
}
}
}
int main()
{
int i, j, k, sum, n, cas = 0;
while (scanf("%s", ch) != EOF)
{
init();
scanf("%d", &n);
for (i = 1;i <= n;i++)
{
scanf("%d%s", &ord[i], str);
id[i] = insert(str);
}
find_fail();
find(ch);
printf("Case %d\n", ++cas);
for (i = 1;i <= n;i++)
printf("%d\n", ans[id[i]][ord[i]]);
printf("\n");
}
return 0;
}