A. Gudako and Ritsuka
留坑。
B. Call of Accepted
留坑。
C. Convex Hull
留坑。
D. Made In Heaven
题意：找第k短路
思路：A_star
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = 1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 10 11 #define N 1005 12 #define M 50005 13 14 int n, m, k; 15 ll T; 16 bool vis[N]; 17 int d[N]; 18 19 struct node { 20 int v; int cost; 21 node() {} 22 node(int v, ll cost) : v(v), cost(cost){} 23 inline bool operator < (const node &b) const { 24 return cost + d[v] > b.cost + d[b.v]; 25 } 26 }edge[M],revedge[M]; 27 28 struct Edge{ 29 int v, cost; 30 inline Edge(){} 31 inline Edge(int v, int cost):v(v), cost(cost){} 32 }; 33 34 vector<Edge>E[N], revE[N]; 35 36 37 inline void Dijstra(int st) 38 { 39 memset(vis, false, sizeof vis); 40 for (int i = 1; i <= n; ++i) 41 { 42 d[i] = INF; 43 } 44 priority_queue<node>q; 45 d[st] = 0; 46 q.push(node(st, 0)); 47 while (!q.empty()) 48 { 49 node tmp = q.top(); 50 q.pop(); 51 int u = tmp.v; 52 if (vis[u]) continue; 53 vis[u] = true; 54 int len = E[u].size(); 55 for (int i = 0; i < len; ++i) 56 { 57 int v = E[u][i].v; 58 int cost = E[u][i].cost; 59 if (!vis[v] && d[v] > d[u] + cost) 60 { 61 d[v] = d[u] + cost; 62 q.push(node(v, d[v])); 63 } 64 } 65 } 66 } 67 68 inline int A_star(int st, int ed) 69 { 70 priority_queue<node>q; 71 q.push(node(st, 0)); 72 --k; 73 while (!q.empty()) 74 { 75 node tmp = q.top(); 76 q.pop(); 77 int u = tmp.v; 78 if (u == ed) 79 { 80 if (k) --k; 81 else return tmp.cost; 82 } 83 int len = revE[u].size(); 84 for (int i = 0; i < len; ++i) 85 { 86 int v = revE[u][i].v; 87 int cost = revE[u][i].cost; 88 q.push(node(v, tmp.cost + cost)); 89 } 90 } 91 return -1; 92 } 93 94 inline void addedge(int u, int v, int w) 95 { 96 revE[u].push_back(Edge(v, w)); 97 E[v].push_back(Edge(u, w)); 98 } 99 100 int st, ed; 101 102 inline void RUN() 103 { 104 while (~scanf("%d %d", &n, &m)) 105 { 106 for (int i = 0; i <= n; ++i) 107 { 108 E[i].clear(); 109 revE[i].clear(); 110 } 111 scanf("%d %d %d %lld", &st, &ed, &k, &T); 112 for (int i = 1; i <= m; ++i) 113 { 114 int u, v, w; 115 scanf("%d %d %d", &u, &v, &w); 116 addedge(u, v, w); 117 } 118 Dijstra(ed); 119 if (d[st] == INF) 120 { 121 puts("Whitesnake!"); 122 continue; 123 } 124 int ans = A_star(st, ed); 125 if (ans == -1) 126 { 127 puts("Whitesnake!"); 128 continue; 129 } 130 puts(ans <= T ? "yareyaredawa" : "Whitesnake!"); 131 } 132 } 133 134 int main() 135 { 136 #ifdef LOCAL_JUDGE 137 freopen("Text.txt", "r", stdin); 138 #endif // LOCAL_JUDGE 139 140 RUN(); 141 142 #ifdef LOCAL_JUDGE 143 fclose(stdin); 144 #endif // LOCAL_JUDGE 145 146 }
E. The cake is a lie
留坑。
F. Fantastic Graph
题意：给出一张二分图，若干条边，选择一些边进去，使得所有点的度数在$[L,R]$ 之间
思路：XDL说爆搜，加上优秀的剪枝(8ms)
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = 1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const int maxn = 1e4 + 10; 10 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 11 12 struct node { 13 int l, r; 14 inline node(){} 15 inline node(int l,int r):l(l),r(r){} 16 }arr[maxn]; 17 18 bool flag = false; 19 int n, m, k; 20 int sum; 21 int res_L; 22 int res_R; 23 int L, R; 24 int du_left[maxn]; 25 int du_right[maxn]; 26 27 inline void Init() 28 { 29 memset(du_left, 0, sizeof du_left); 30 memset(du_right, 0, sizeof du_right); 31 } 32 33 inline void DFS(int idx, int cnt) 34 { 35 if (sum == n + m) 36 { 37 flag = true; 38 return; 39 } 40 if (idx > k) return; 41 if (k - idx + 1 + cnt < res_L) return; 42 if (cnt >= res_R) return; 43 if (flag) return; 44 //do 45 if (du_left[arr[idx].l] < R && du_right[arr[idx].r] < R) 46 { 47 du_left[arr[idx].l]++; 48 du_right[arr[idx].r]++; 49 if (du_left[arr[idx].l] == L) sum++; 50 if (du_right[arr[idx].r] == L) sum++; 51 DFS(idx + 1, cnt + 1); 52 if (flag) return; 53 if (du_left[arr[idx].l] == L) sum--; 54 if (du_right[arr[idx].r] == L) sum--; 55 du_left[arr[idx].l]--; 56 du_right[arr[idx].r]--; 57 } 58 //not do 59 DFS(idx + 1, cnt); 60 if (flag) return; 61 } 62 63 inline void RUN() 64 { 65 int cas = 0; 66 while (~scanf("%d %d %d", &n, &m, &k)) 67 { 68 printf("Case %d: ", ++cas); 69 Init(); 70 scanf("%d %d", &L, &R); 71 int cnt = 0; 72 for (int i = 1; i <= k; ++i) 73 { 74 scanf("%d %d", &arr[i].l, &arr[i].r); 75 du_left[arr[i].l]++; 76 du_right[arr[i].r]++; 77 if (du_left[arr[i].l] == L) cnt++; 78 if (du_right[arr[i].r] == L) cnt++; 79 } 80 if (L == 0) 81 { 82 puts("Yes"); 83 continue; 84 } 85 if (cnt != n + m ) 86 { 87 puts("No"); 88 continue; 89 } 90 sum = 0; 91 res_L = ((n + m) * L + 1) / 2; 92 res_R = ((n + m) * R) / 2; 93 Init(); 94 flag = false; 95 DFS(1, 0); 96 puts(flag ? "Yes" : "No"); 97 } 98 } 99 100 int main() 101 { 102 #ifdef LOCAL_JUDGE 103 freopen("Text.txt", "r", stdin); 104 #endif // LOCAL_JUDGE 105 106 RUN(); 107 108 #ifdef LOCAL_JUDGE 109 fclose(stdin); 110 #endif // LOCAL_JUDGE 111 112 }
G. Spare Tire
题意：
思路：
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = 1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const int maxn = 1e4 + 10; 10 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 11 12 int tot; 13 bool isprime[maxn]; 14 int prime[maxn]; 15 ll inv[maxn]; 16 17 18 inline void Init_prime() 19 { 20 inv[1] = 1; 21 for (int i = 2; i < maxn; ++i) 22 { 23 inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD; 24 } 25 memset(isprime, true, sizeof isprime); 26 tot = 0; 27 for (int i = 2; i < maxn; ++i) 28 { 29 if (isprime[i]) 30 { 31 prime[tot++] = i; 32 for (int j = (i << 1); j < maxn; j += i) 33 { 34 isprime[i] = false; 35 } 36 } 37 } 38 } 39 40 vector<ll>vec; 41 ll n, m; 42 ll ans; 43 44 inline void Init(ll x) 45 { 46 vec.clear(); 47 for (int i = 0; i < tot && prime[i] < x; ++i) 48 { 49 if (x % prime[i] == 0) 50 { 51 vec.push_back(prime[i]); 52 while (x % prime[i] == 0) 53 { 54 x /= prime[i]; 55 } 56 } 57 } 58 if (x != 1) vec.push_back(x); 59 } 60 61 inline void RUN() 62 { 63 Init_prime(); 64 while (~scanf("%lld %lld", &n, &m)) 65 { 66 ans = n % MOD * (n + 1) % MOD * (n + 2) % MOD * inv[3] % MOD; 67 if (m == 1) 68 { 69 printf("%lld\n", ans); 70 continue; 71 } 72 Init(m); 73 int len = vec.size(); 74 for (int i = 1; i < (1 << len); ++i) 75 { 76 int cnt = 0; 77 ll t = 1; 78 for (int j = 0; j < len; ++j) 79 { 80 if (i & (1 << j)) 81 { 82 cnt++; 83 t *= vec[j]; 84 } 85 } 86 ll x = n / t; 87 ll tmp = x % MOD * (x + 1) % MOD * (2 * x % MOD + 1) % MOD * inv[6] % MOD * t % MOD * (t + 1) % MOD - x % MOD * (x + 1) % MOD * (x - 1) % MOD * inv[3] % MOD * t % MOD; 88 tmp = (tmp + MOD) % MOD; 89 if (cnt & 1) 90 { 91 ans = (ans - tmp + MOD) % MOD; 92 } 93 else 94 { 95 ans = (ans + tmp) % MOD; 96 } 97 } 98 printf("%lld\n", ans); 99 } 100 } 101 102 int main() 103 { 104 #ifdef LOCAL_JUDGE 105 freopen("Text.txt", "r", stdin); 106 #endif // LOCAL_JUDGE 107 108 RUN(); 109 110 #ifdef LOCAL_JUDGE 111 fclose(stdin); 112 #endif // LOCAL_JUDGE 113 114 }
H. Hamming Weight
留坑。
I. Lattice's basics in digital electronics
按题意模拟即可。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 6 int t, n, m; 7 string str, tmp, ttmp, ans, s; 8 unordered_map <string, char> mp; 9 unordered_map <char, string> mmp; 10 11 inline void Init() 12 { 13 mmp.clear(); 14 mmp['0'] = "0000"; 15 mmp['1'] = "0001"; 16 mmp['2'] = "0010"; 17 mmp['3'] = "0011"; 18 mmp['4'] = "0100"; 19 mmp['5'] = "0101"; 20 mmp['6'] = "0110"; 21 mmp['7'] = "0111"; 22 mmp['8'] = "1000"; 23 mmp['9'] = "1001"; 24 mmp['A'] = "1010"; 25 mmp['B'] = "1011"; 26 mmp['C'] = "1100"; 27 mmp['D'] = "1101"; 28 mmp['E'] = "1110"; 29 mmp['F'] = "1111"; 30 mmp['a'] = "1010"; 31 mmp['b'] = "1011"; 32 mmp['c'] = "1100"; 33 mmp['d'] = "1101"; 34 mmp['e'] = "1110"; 35 mmp['f'] = "1111"; 36 } 37 38 inline bool ok(string s) 39 { 40 int cnt = 0; 41 for (int i = 0; i < 8; ++i) cnt += (s[i] == '1'); 42 if ((s[8] == '0' && (cnt & 1)) || (s[8] == '1' && (cnt & 1) == 0)) return true; 43 return false; 44 } 45 46 inline void Run() 47 { 48 cin.tie(0), cout.tie(0); Init(); 49 ios::sync_with_stdio(false); 50 cin >> t; 51 while (t--) 52 { 53 mp.clear(); cin >> m >> n; 54 for (int i = 1, num; i <= n; ++i) 55 { 56 cin >> num >> str; 57 mp[str] = num; 58 } 59 cin >> s; tmp = ""; str = ""; 60 for (int i = 0, len = s.size(); i < len; ++i) 61 tmp += mmp[s[i]]; 62 for (int i = 0, len = tmp.size(); i < len; ) 63 { 64 if (len - i < 9) break; 65 ttmp = ""; 66 for (int cnt = 0; cnt < 9; ++cnt, ++i) 67 ttmp += tmp[i]; 68 if (ok(ttmp)) 69 { 70 ttmp.erase(ttmp.end() - 1); 71 str += ttmp; 72 //cout << ttmp << endl; 73 } 74 } 75 //for (auto it : mp) cout << it.first << " " << it.second << endl; 76 ans.clear(), tmp.clear(); 77 for (int i = 0, len = str.size(); i < len && ans.size() < m; ++i) 78 { 79 tmp += str[i]; 80 if (mp.find(tmp) != mp.end()) 81 { 82 ans += mp[tmp]; 83 tmp = ""; 84 } 85 } 86 cout << ans << endl; 87 } 88 } 89 90 int main() 91 { 92 #ifdef LOCAL 93 freopen("Test.in", "r", stdin); 94 #endif 95 96 Run(); 97 return 0; 98 }
J. Ka Chang
题意：两种操作 $1 L X$ 所有深度为L的加上x $2 X$ 查询以x为根的所有子节点的和
思路：以x为根的子节点的和可以用DFS序使得所有子树的标号在一块
然后对于更新操作，考虑两种方法操作
1° 暴力更新每个点
2° 记录这个深度更新了多少，查询的时候找出这个深度中有子节点有几个 直接加上
如果只用第二个操作更新，那么当给的树是一条链的时候，查询的复杂度达到$O(nlogn)$
只用第一个操作更新，那么更新的操作当给的树是菊花形的时候，更新的复杂度达到$O(nlogn)$
那么考虑将两种操作结合，当某个深度中元素个数大于sqrt(100000) 的时候，用第二种操作 否则用第一种
然后查询的时候 要记得 都加上
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define ll long long 6 7 struct Edge 8 { 9 int to, nx; 10 inline Edge() {} 11 inline Edge(int to, int nx) : to(to), nx(nx) {} 12 }edge[N << 1]; 13 14 int n, q, Maxdeep; 15 int head[N], pos; 16 int ord[N], son[N], deep[N], fa[N], cnt; 17 vector <int> dep[N], Lar; 18 ll sum[N]; 19 20 inline void Init() 21 { 22 memset(head, -1, sizeof head); 23 pos = 0; cnt = 0; Maxdeep = 0; Lar.clear(); 24 for (int i = 0; i <= 1; ++i) dep[i].clear(); 25 fa[1] = 1; deep[1] = 0; 26 } 27 28 inline void addedge(int u, int v) 29 { 30 edge[++pos] = Edge(v, head[u]); head[u] = pos; 31 } 32 33 inline void DFS(int u, int pre) 34 { 35 ord[u] = ++cnt; 36 dep[deep[u]].emplace_back(cnt); 37 Maxdeep = max(Maxdeep, deep[u]); 38 for (int it = head[u]; ~it; it = edge[it].nx) 39 { 40 int v = edge[it].to; 41 if (v == pre) continue; 42 deep[v] = deep[u] + 1; 43 DFS(v, u); 44 } 45 son[u] = cnt; 46 } 47 48 struct node 49 { 50 int l, r; 51 ll sum; 52 inline node() {} 53 inline node(int l, int r, ll sum) : l(l), r(r), sum(sum) {} 54 }tree[N << 2]; 55 56 inline void pushup(int id) 57 { 58 tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum; 59 } 60 61 inline void build(int id, int l, int r) 62 { 63 tree[id] = node(l, r, 0); 64 if (l == r) return; 65 int mid = (l + r) >> 1; 66 build(id << 1, l, mid); 67 build(id << 1 | 1, mid + 1, r); 68 } 69 70 inline void update(int id, int pos, ll val) 71 { 72 if (tree[id].l == tree[id].r) 73 { 74 tree[id].sum += val; 75 return; 76 } 77 int mid = (tree[id].l + tree[id].r) >> 1; 78 if (pos <= mid) update(id << 1, pos, val); 79 else if (pos > mid) update(id << 1 | 1, pos, val); 80 pushup(id); 81 } 82 83 ll anssum; 84 85 inline void query(int id, int l, int r) 86 { 87 if (tree[id].l >= l && tree[id].r <= r) 88 { 89 anssum += tree[id].sum; 90 return; 91 } 92 int mid = (tree[id].l + tree[id].r) >> 1; 93 if (l <= mid) query(id << 1, l, r); 94 if (r > mid) query(id << 1 | 1, l, r); 95 } 96 97 inline void Run() 98 { 99 while (scanf("%d%d", &n, &q) != EOF) 100 { 101 Init(); 102 for (int i = 1, u, v; i < n; ++i) 103 { 104 scanf("%d%d", &u, &v); 105 addedge(u, v); addedge(v, u); 106 } 107 DFS(1, 1); build(1, 1, n); 108 int limit = (int)sqrt(100000); 109 for (int i = 1; i <= Maxdeep; ++i) 110 { 111 if (dep[i].size() >= limit) 112 { 113 Lar.emplace_back(i); 114 sort(dep[i].begin(), dep[i].end()); 115 } 116 } 117 for (int i = 1, op, l, x; i <= q; ++i) 118 { 119 scanf("%d", &op); 120 if (op == 1) 121 { 122 scanf("%d%d", &l, &x); 123 if (dep[l].size() < limit) 124 { 125 for (auto it : dep[l]) update(1, it, (ll)x); 126 } 127 else 128 sum[l] += (ll)x; 129 } 130 else 131 { 132 scanf("%d", &x); 133 anssum = 0; query(1, ord[x], son[x]); 134 ll ans = anssum; 135 int l = ord[x], r = son[x]; 136 for (auto it : Lar) 137 { 138 if (it < deep[x]) continue; 139 int k = upper_bound(dep[it].begin(), dep[it].end(), r) - lower_bound(dep[it].begin(), dep[it].end(), l); 140 if (k == 0) break; 141 ans += sum[it] * k; 142 } 143 printf("%lld\n", ans); 144 } 145 } 146 } 147 } 148 149 int main() 150 { 151 #ifdef LOCAL 152 freopen("Test.in", "r", stdin); 153 #endif 154 155 Run(); 156 return 0; 157 }
K. Supreme Number
只有几个数，爆搜出来，判断一下
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const int MOD = 1e9 + 7; 8 const int INF = 0x3f3f3f3f; 9 const int maxn = 2e6 + 10; 10 11 string s[] = { "1","2","3","5","7","11","13","17","23","31","37","53","71","73","113","131","137","173","311","317" }; 12 13 inline void RUN() 14 { 15 int t; 16 cin >> t; 17 for (int cas = 1; cas <= t; ++cas) 18 { 19 cout << "Case #" << cas << ": "; 20 string ans; 21 string str; 22 cin >> str; 23 for (int i = 0; i < 20; ++i) 24 { 25 if (str.length() == s[i].length()) 26 { 27 if (str >= s[i]) ans = s[i]; 28 } 29 else if (str.length() < s[i].length()) 30 { 31 continue; 32 } 33 else ans = s[i]; 34 } 35 cout << ans << endl; 36 } 37 } 38 39 int main() 40 { 41 #ifdef LOCAL_JUDGE 42 freopen("Text.txt", "r", stdin); 43 #endif // LOCAL_JUDGE 44 ios::sync_with_stdio(false); 45 cin.tie(0); 46 cout.tie(0); 47 RUN(); 48 49 #ifdef LOCAL_JUDGE 50 fclose(stdin); 51 #endif // LOCAL_JUDGE 52 53 }