Codeforces Round #457 (Div. 2)
| Codeforces Round #457 (Div. 2) |
|---|
A.模拟。
B.模拟。比赛的时候这题出锅了，然后就unrated了。注意要先保证最大值最小，再保证字典序最大。
C.构造。挑一个大于N的质数然后连啊连就可以了。
D.可持久化Trie。弄两个Trie。一个记录字符串，另一个记录数字。
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 struct StringTrie { 6 StringTrie *ch[26]; int val; 7 StringTrie() {memset(this, 0, sizeof(*this)); val = -1;} 8 StringTrie* set(int n, const char *str, int p, int v) { 9 StringTrie *res = new StringTrie(); *res = *this; 10 if (p >= n) res->val = v; 11 else { 12 int c = str[p] - 'a'; 13 if (!ch[c]) ch[c] = new StringTrie(); 14 res->ch[c] = ch[c]->set(n, str, p + 1, v); 15 } return res; 16 } 17 int query(int n, const char *str, int p) { 18 if (p >= n) return val; 19 else { 20 int c = str[p] - 'a'; 21 if (!ch[c]) return -1; 22 return ch[c]->query(n, str, p + 1); 23 } 24 } 25 } *srt[100005]; 26 struct BinaryTrie { 27 BinaryTrie *ch[2]; int val; 28 BinaryTrie() {memset(this, 0, sizeof(*this));} 29 BinaryTrie* add(int x, int val, int p) { 30 BinaryTrie *res = new BinaryTrie(); *res = *this; res->val += val; 31 if (p >= 0) { 32 int c = (x >> p) & 1; 33 if (!ch[c]) ch[c] = new BinaryTrie(); 34 res->ch[c] = ch[c]->add(x, val, p - 1); 35 } return res; 36 } 37 int query(int x, int p) { 38 if (p < 0) return 0; 39 else { 40 int c = (x >> p) & 1, res = 0; 41 if (c && ch[0]) res += ch[0]->val; 42 if (ch[c]) res += ch[c]->query(x, p - 1); 43 return res; 44 } 45 } 46 } *brt[100005]; 47 int main() { 48 srt[0] = new StringTrie(); brt[0] = new BinaryTrie(); 49 int n; scanf("%d", &n); char op[20], str[20]; 50 for (int v, i = 1; i <= n; ++i) { 51 scanf("%s", op); 52 if (*op == 's') { 53 scanf("%s%d", str, &v); int l = strlen(str); 54 int x = srt[i-1]->query(l, str, 0); 55 srt[i] = srt[i-1]->set(l, str, 0, v); 56 brt[i] = brt[i-1]; 57 if (~x) brt[i] = brt[i]->add(x, -1, 30); 58 brt[i] = brt[i]->add(v, 1, 30); 59 } else if (*op == 'r') { 60 scanf("%s", str); int l = strlen(str); 61 int x = srt[i-1]->query(l, str, 0); 62 brt[i] = brt[i-1]; srt[i] = srt[i-1]->set(l, str, 0, -1); 63 if (~x) brt[i] = brt[i]->add(x, -1, 30); 64 } else if (*op == 'q') { 65 scanf("%s", str); int l = strlen(str); 66 brt[i] = brt[i-1]; srt[i] = srt[i-1]; 67 int x = srt[i]->query(l, str, 0); 68 if (~x) printf("%d\n", brt[i]->query(x, 30)); 69 else puts("-1"); fflush(stdout); 70 } else { 71 scanf("%d", &v); srt[i] = srt[i-v-1]; brt[i] = brt[i-v-1]; 72 } 73 } 74 return 0; 75 }
E.跟bzoj上的遥远的国度很类似。
