Contact

这道题用trie树来做。我做的真是太悲剧了，在建树的时候，上界我用的strlen(ss)，结果光建树就花了8秒。刚开始我以为是因为指针建树花时间太多，全改成数组后，还是一样，我才改的strlen(ss)，结果速度一下子就提上来了，真是无语……

 注意输出格式，每行最多输出6个。

程序代码：

/*ID:zhaofuk1PROG: contactLANG: C++*/#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <vector>#include <utility>#include <algorithm>using namespace std;#define MAXN 100000typedef struct TrieNode{ int nCount; struct TrieNode *next[2]; void init(){ nCount = 1; next[0] = next[1] = NULL; }}TrieNode;TrieNode memory[MAXN];TrieNode *nRoot;int allocp, lower, upper;typedef pair<int, string> pill;vector<pill> vec;int cmp(pill a, pill b){ if(a.first == b.first){ if(a.second.size() == b.second.size()) return a.second < b.second; else return a.second.size() < b.second.size(); }else return a.first > b.first;}void nInit(){ allocp = 0; nRoot = &memory[allocp++]; nRoot->init();}void InsertTrie(char *s, int len){ TrieNode *p = nRoot; int i, k; for(i = 0, k; i < len && i < upper; i++){ k = s[i] - '0'; if(!p->next[k]){ p->next[k] = &memory[allocp++]; p->next[k]->init(); }else p->next[k]->nCount++; p = p->next[k]; }}int SearchTrie(char *s){ TrieNode *p = nRoot; for(int i = 0, k; i < strlen(s); i++){ k = s[i] - '0'; if(!p->next[k]) return 0; else p = p->next[k]; } return p->nCount;}void dfs(int cur, int n, char *s){ if(cur == n){ s[cur] = '/0'; int t; if(t = SearchTrie(s)){ pill u; u.first = t; u.second.assign(s); vec.push_back(u); } return ; } s[cur] = '0'; dfs(cur + 1, n, s); s[cur] = '1'; dfs(cur + 1, n, s);}int main(){ int n, bit, t, len; char ss[2 * MAXN + 10], s[20]; char tt[100]; nInit(); freopen("contact.in", "r", stdin); freopen("contact.out", "w", stdout); scanf("%d %d %d", &lower, &upper, &n); scanf("%s", ss); while(scanf("%s", tt) == 1){ strcat(ss, tt); } len = strlen(ss); for(int i = 0; i < len; i++){ InsertTrie(ss + i, len - i); } for(int i = lower; i <= upper; i++) dfs(0, i, s); sort(vec.begin(), vec.end(), cmp); if(!vec.empty()){ cout<<vec[0].first<<endl<<vec[0].second; t = vec[0].first; } int i, j, k; for(i = 1, j = 0, k = 0; j < n && i < vec.size(); i++){ if(vec[i].first == t){ k++; if(k == 6){ cout<<endl<<vec[i].second; k = 0; }else cout<<" "<<vec[i].second; }else{ cout<<endl; j++; k = 0; t = vec[i].first; if(j < n) cout<<vec[i].first<<endl<<vec[i].second; } } if(j < n) cout<<endl; return 0;}