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  • hdu 2846 Repository

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=2846  

    Repository

    Description

    When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.

    Input

    There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.

    Output

    For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

    Sample Input

    20
    ad
    ae
    af
    ag
    ah
    ai
    aj
    ak
    al
    ads
    add
    ade
    adf
    adg
    adh
    adi
    adj
    adk
    adl
    aes
    5
    b
    a
    d
    ad
    s

    Sample Output

    0
    20
    11
    11
    2

    字典树,g++ mle c++ ac。。

    #include<algorithm>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<set>
    using std::set;
    using std::min;
    using std::find;
    using std::pair;
    using std::vector;
    using std::multiset;
    #define pb(e) push_back(e)
    #define sz(c) (int)(c).size()
    #define mp(a, b) make_pair(a, b)
    #define all(c) (c).begin(), (c).end()
    #define iter(c) __typeof((c).begin())
    #define cls(arr, val) memset(arr, val, sizeof(arr))
    #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    #define rep(i, n) for(int i = 0; i < (int)n; i++)
    #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
    const int N = 100010;
    const int INF = 0x3f3f3f3f;
    struct Node {
    	Node *ch[26];
    	int cnt;
    	int f;
    	Node() {
    		rep(i, 26) ch[i] = NULL;
    		cnt = 0;
    		f = -1;
    	}
    }*root;
    void insert(Node *x, char *str, int f) {
    	int d = 0;
    	char *p = str;
    	while (*p != '') {
    		d = *p - 'a';
    		if (!x->ch[d]) x->ch[d] = new Node();
    		x = x->ch[d];
    		if (x->f != f) {
    			x->f = f;
    			x->cnt++;
    		}
    		p++;
    	}
    }
    int query(Node *x, char *str) {
    	char *p = str;
    	while (*p != '') {
    		if (!x || !x->ch[*p - 'a']) return 0;
    		x = x->ch[*p - 'a'];
    		p++;
    	}
    	return x->cnt;
    }
    void built(int t) {
    	int n;
    	char buf[100];
    	rep(i, t) {
    		scanf("%s", buf);
    		n = strlen(buf);
    		rep(j, n) {
    			insert(root, buf + j, i);
    		}
    	}
    }
    void solve(int n) {
    	int m;
    	char buf[100];
    	root = new Node();
    	built(n);
    	scanf("%d", &m);
    	while (m--) {
    		scanf("%s", buf);
    		printf("%d
    ", query(root, buf));
    	}
    }
    int main() {
    #ifdef LOCAL
    	freopen("in.txt", "r", stdin);
    	freopen("out.txt", "w+", stdout);
    #endif
    	int n;
    	while (~scanf("%d", &n)) {
    		solve(n);
    	}
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GadyPu/p/4792900.html
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