ZOJ 3674 Search in the Wiki（字典树 + map + vector）

来源：互联网发布：知乎上面的神回复编辑：程序博客网时间：2024/05/01 18:17

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4917

题意：每个单词都一些tips单词。先输入n个单词和他们的tips。然后m组查询，每次查询一些单词，按字典序输出这些单词的公有tips。（每个单词都都只包含小写大写字母）

思路：对第i个单词，用vector数组g，g[i]来存这个单词的所有tips。对于所有单词建立字典树，在单词的结尾结点存好该单词的tips在g数组中存的一维下标i。最后用map来计数每组询问中每个tips。每组询问中，对于所有的查询单词中所有的tips计数值等于单词个数的进行记录，最终按字典序排序，最后输出。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <map>#include <set>#include <algorithm>#include <vector>using namespace std;const int N = 3e3 + 10;const int SIZE = 60;const int MAX_WORD = 305;struct Trie {int val[SIZE];int w;};int sz;char str[MAX_WORD];char st[MAX_WORD];Trie pn[N];map<string, int> m;vector<string> g[N];vector<string> res;int newnode() {memset(pn[sz].val, 0, sizeof(pn[sz].val));pn[sz].w = -1;return sz++;}void init() {sz = 0;newnode();}void insert(char *s, int j) {int u = 0;int len = strlen(s);for (int i = 0; i < len; i++) {int idx = s[i] - 'A';if (!pn[u].val[idx])pn[u].val[idx] = newnode();u = pn[u].val[idx];}pn[u].w = j;string t;gets(str);len = strlen(str);for (int i = 0; i < len; i++) {if (str[i] == ' ') {g[j].push_back(t);t.clear();}elset.push_back(str[i]);}if (!t.empty())g[j].push_back(t);}int findstr(char *s) {int u = 0;int len = strlen(s);for (int i = 0; i < len; i++) {int idx = s[i] - 'A';if (!pn[u].val[idx])return -1;u = pn[u].val[idx];}if (g[pn[u].w].empty())return -1;for (int i = 0; i < g[pn[u].w].size(); i++)m[g[pn[u].w][i]]++;return u;}int main() {int n, q;while (scanf("%d", &n) != EOF) {init();for (int i = 0; i < n; i++) {scanf("%s", str);getchar();g[i].clear();insert(str, i);}scanf("%d", &q);getchar();int k, t;for (int i_q = 1; i_q <= q; i_q++) {m.clear();gets(str);int y = 0;t = 0;int len = strlen(str);for (int i = 0; i < len; i++) {st[y++] = str[i];if (str[i + 1] == '\0' || str[i + 1] == ' ') {st[y] = '\0';k = findstr(st);if (k == -1)break;t++;y = 0;i++;}}if (k == -1) {puts("NO");continue;}res.clear();for (int i = 0; i < g[pn[k].w].size(); i++)if (m[g[pn[k].w][i]] == t)res.push_back(g[pn[k].w][i]);if (res.empty())puts("NO");else {sort(res.begin(), res.end());for (int i = 0; i < res.size() - 1; i++)cout << res[i] << " ";cout << res[res.size() - 1] << endl;}}}return 0;}

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