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18000 Two String

时间限制:1000MS  内存限制:65535K
提交次数:0 通过次数:0

题型: 编程题   语言: 不限定

Description

Given two string A and B and three kinds of operations as following:
(1)Push_back c            add a character c at the back of string B
(2)Push_front c           add a character c in front of string B
(3)Query                  calculate and output how many times B appears in A
could you calculate and output the right answer for each query?

输入格式

The first line contains the string A ,the second line contains the string B ,the third line contains an integer M (1 <= M <= 2000), expressing the number of operation, Each of the following M lines contains one of the above-mentioned three operations.
total length of the string does not exceed 5000,all character in the input are the lower case Latin alphabet

输出格式

For each query, output a line containing the answer.

输入样例

abcabc
a
5
Query
Push_back b
Query
Push_front a
Query

aaaaa
a
5
Query
Push_back a
Query
Push_front a
Query

输出样例

2
2
0

5
4
3

来源

星尘

作者

admin

一定要注意到的是，

total length of the string does not exceed 5000,

就是所有样例的字符全加起来不会超过5000，其实我觉得这样给数据范围很坑爹。不如一个样例一个样例给我。

一直不敢做，其实就是暴力。

对于每种push，暴力进行。

每种查询，kmp一次。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 10000 + 20;
char str[maxn];
char sub[2][maxn];
int lensub;
int lenstr;
int now;
int tonext[maxn];
void get_next(int now) {
    int i = 1, j = 0;
    tonext[1] = 0;
    while (i <= lensub) {
        if (j == 0 || sub[now][i] == sub[now][j]) {
            tonext[++i] = ++j;
        } else j = tonext[j];
    }
}
int kmp(int now) {
    get_next(now);
    int i = 1, j = 1;
    int ans = 0;
    while (i <= lenstr) {
        if (j == 0 || str[i] == sub[now][j]) {
            ++i;
            ++j;
        } else j = tonext[j];
        if (j == lensub + 1) {
            ans++;
            j = tonext[j];
        }
    }
    return ans;
}
void work() {
    lenstr = strlen(str + 1);
    lensub = strlen(sub[now] + 1);
    int q;
    scanf("%d", &q);
    char t[22];
    for (int i = 1; i <= q; ++i) {
        scanf("%s", t + 1);
        if (t[1] == ‘Q‘) {
            printf("%d\n", kmp(now));
        } else {
            char tt[23];
            scanf("%s", tt);
            if (strcmp("Push_back", t + 1) == 0) {
                sub[now][++lensub] = tt[0];
            } else {
                for (int i = 1; i <= lensub; ++i) {
                    sub[!now][i + 1] = sub[now][i];
                }
                sub[!now][1] = tt[0];
                now = !now;
                lensub += 1;
            }
        }
    }
}
int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    while (scanf("%s%s", str + 1, sub[now] + 1) != EOF) work();
    return 0;
}