题意:询问所有字符串的比较次数和(注意for循环内的比较也算)
分析:将所有字符串插入到字典树上,然后结点信息记录有几个字符串,那么每走到一个结点就能知道比较到此时需要的次数。学习到链表存结点
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 4e3 + 5; const int M = 1e3 + 5; const int NODE = N * M; struct Trie { int head[NODE], nex[NODE]; char ch[NODE]; int cnt[NODE]; int sz; void clear() { sz = 1; cnt[0] = head[0] = nex[0] = 0; } void insert(char *str) { int u = 0, len = strlen (str); cnt[0]++; for (int i=0; i<=len; ++i) { bool found = false; int v; for (v=head[u]; v; v=nex[v]) { if (ch[v] == str[i]) { found = true; break; } } if (!found) { //插入到链表的头结点后一个 v = sz++; cnt[v] = 0; ch[v] = str[i]; nex[v] = head[u]; head[u] = v; head[v] = 0; } u = v; cnt[u]++; } } void query(int u, int dep, ll &res) { if (!head[u]) { res += cnt[u] * (cnt[u] - 1) * dep; return ; } int sum = 0, tmp = cnt[u]; for (int v=head[u]; v; v=nex[v]) { //sum += cnt[v] * (cnt[u] - cnt[v]); res += cnt[v] * (tmp - cnt[v]) * (dep * 2 + 1); tmp -= cnt[v]; } //res += sum / 2 * (2 * dep + 1); for (int v=head[u]; v; v=nex[v]) { query (v, dep + 1, res); } } }trie; char word[M]; int main(void) { int n, cas = 0; while (scanf ("%d", &n) == 1) { if (!n) break; trie.clear (); for (int i=0; i<n; ++i) { scanf ("%s", &word); trie.insert (word); } ll ans = 0; trie.query (0, 0, ans); printf ("Case %d: %lld ", ++cas, ans); } return 0; }