Weekly Challenges - Week 11
一拖再拖，忍无可忍，自己懒的没救了。
一周前就结束的比赛，到现在才想起来补。
最后一题貌似又遇到了splay。。。还是不会！！！shit
Ａ题－－快速幂
1 #include <cmath> 2 #include <cstdio> 3 #include <vector> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 const int MOD = 1e9 + 7; 9 10 int mod_pow(int a, int b) { 11 int res = 1; 12 while (b) { 13 if (b & 1) res = (res * (long long)a) % MOD; 14 a = (a * (long long)a) % MOD; 15 b >>= 1; 16 } 17 return res; 18 } 19 20 int main() { 21 /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 22 int T; 23 ios::sync_with_stdio(false); 24 cin >> T; 25 while (T--) { 26 int t; 27 cin >> t; 28 cout << (2 + mod_pow(2, t + 1)) % MOD << endl; 29 } 30 return 0; 31 }
Ｂ题
如果说这题的困难之处，那就是要看出来其实这种数的个数很少。所以先把所有的范围之内的数打表出来。
1 #include <cmath> 2 #include <cstdio> 3 #include <vector> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 typedef long long LL; 9 #define rep(i, n) for (int i = (1); i <= (n); i++) 10 LL pw10[20]; 11 const LL MAX_N = (LL)1e18; 12 13 int main() { 14 pw10[0] = 1LL; 15 rep (i, 19) pw10[i] = pw10[i - 1] * 10; 16 vector<LL> ans; 17 rep (i, 9) ans.push_back(i); 18 int cur = 0, cur_len = 2, next = -1; 19 while (true) { 20 if (next == - 1 && ans[cur] * (cur_len + 1) >= pw10[cur_len]) { 21 next = cur; 22 } 23 LL x = ans[cur] * cur_len; 24 if (x >= MAX_N) break; 25 if (pw10[cur_len - 1] <= x && x < pw10[cur_len]) { 26 ans.push_back(x); 27 } else if (x >= pw10[cur_len]) { 28 cur = next; 29 next = -1; 30 cur_len++; 31 continue; 32 } 33 cur++; 34 } 35 //rep (i, 25) cerr << ans[i] << endl; 36 int T; 37 cin >> T; 38 while (T--) { 39 LL A, B; 40 cin >> A >> B; 41 LL rgt = upper_bound(ans.begin(), ans.end(), B) - ans.begin(); 42 LL lft = lower_bound(ans.begin(), ans.end(), A) - ans.begin(); 43 LL res = rgt - lft; 44 if (A == 0) res++; 45 cout << res << endl; 46 } 47 48 return 0; 49 }
Ｃ题
题意：给出Ｎ个结点的树，每个结点有一个权值，在树上有一种操作，可以任选一个结点，然后删除以该结点为根的子树。最多执行Ｋ次这种操作
使得树上的权值总和最大。
首先，如果按照前序遍历访问每个结点的顺序给结点做标记，那么以某个结点为根的子树中所有结点的标记一定该结点的后面。
dfs时顺便维护一下以每个结点为根的子树中结点的个数tot[i]。
dp[i][j]表示到dfs中第ｉ个遍历的结点为止操作了ｊ次可以获得的最大权值。
这样当到达第ｉ个结点时，dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + value[i]) //不删除
dp[i + tot[i]][j + 1] = max(dp[i + tot[i]][j + 1], dp[i][j]) ／／删除
1 #include <cmath> 2 #include <cstdio> 3 #include <vector> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 const int MAX_N = 100050; 9 typedef long long LL; 10 const LL INF = (LL)1e19; 11 vector<int> G[MAX_N]; 12 int w[MAX_N], vs[MAX_N], tot[MAX_N], cnt; 13 LL dp[MAX_N][205]; 14 15 void dfs(int u, int fa) { 16 int sz = (int)G[u].size(); 17 vs[++cnt] = u; 18 tot[u] = 1; 19 if (sz == 1) { 20 return ; 21 } 22 for (int i = 0; i < sz; i++) { 23 int v = G[u][i]; 24 if (v == fa) continue; 25 dfs(v, u); 26 tot[u] += tot[v]; 27 } 28 } 29 30 void read(int &res) { 31 res = 0; 32 int sign = 1; 33 char c = ' '; 34 while (c < '0' || c > '9') { 35 if (c == '-') sign = -1; 36 c = getchar(); 37 } 38 while (c >= '0' && c <= '9') res = (res<<3) + (res<<1) + c - '0', c = getchar(); 39 if (sign == -1) res = -res; 40 } 41 42 int main() { 43 ios::sync_with_stdio(false); 44 int N, K; 45 cin >> N >> K; 46 //read(N); read(K); 47 for (int i = 1; i <= N; i++) cin >> w[i];//read(w[i]); 48 for (int i = 1; i < N; i++) { 49 int u, v; 50 cin >> u >> v; 51 //read(u), read(v); 52 G[u].push_back(v); 53 G[v].push_back(u); 54 } 55 cnt = 0; 56 dfs(1, -1); 57 for (int i = 0; i <= K; i++) for (int j = 1; j <= N; j++) dp[j][i] = -INF; 58 //for (int i = 0; i <= K; i++) dp[0][i] = 0; 59 dp[1][0] = 0; 60 for (int i = 1; i <= N; i++) { 61 int u = vs[i]; 62 for (int j = 0; j <= K; j++) if (dp[i][j] != -INF) { 63 dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + w[u]); 64 if (j < K) 65 dp[i + tot[u]][j + 1] = max(dp[i + tot[u]][j + 1], dp[i][j]); 66 } 67 } 68 LL res = -INF; 69 for (int i = 0; i <= K; i++) res = max(res, dp[N + 1][i]); 70 cout << res << endl; 71 72 return 0; 73 }
Ｄ题
题意：Ｎ种技能，第ｉ种技能有ci个人学会，有Ｔ个巫师，每个巫师有两个集合Ａ和Ｂ，他可以把一个拥有Ａ集合中的技能的人转移到集合Ｂ中的某个技能中去。
问最多可以使多少个技能至少有一个人学会。
增加一个源点和汇点，源点到Ｎ种技能连一条容量为ci的边，从Ｎ种技能到汇点连一条容量为１的边。
对于每个巫师，从Ａ集合中的技能连一条容量为１的边到该巫师，从该巫师连一条容量为１的边到Ｂ集合中的每种技能。
最大流即为答案。。。
1 #include <queue> 2 #include <cmath> 3 #include <cstdio> 4 #include <vector> 5 #include <cstring> 6 #include <iostream> 7 #include <algorithm> 8 using namespace std; 9 10 struct edge { 11 int to, cap, rev; 12 edge() {} 13 edge(int _to, int _cap, int _rev):to(_to), cap(_cap), rev(_rev) {} 14 }; 15 const int INF = 0x3f3f3f3f; 16 const int MAX_V = 250; // ????` 17 vector<edge> G[MAX_V]; // ????? 18 int level[MAX_V]; // ?????????? 19 int iter[MAX_V]; // ???????????????? 20 int Q[MAX_V]; // BFS?? 21 22 void add_edge(int from, int to, int cap) { 23 G[from].push_back(edge(to, cap, G[to].size())); 24 G[to].push_back(edge(from, 0, G[from].size() - 1)); 25 } 26 27 // ????????????????? 28 void bfs(int s) { 29 memset(level, -1, sizeof(level)); 30 int ini = 0, end = 0; 31 Q[end++] = s; 32 level[s] = 0; 33 while (end > ini) { 34 int v = Q[ini++]; 35 for (int i = 0; i < G[v].size(); i++) { 36 edge &e = G[v][i]; 37 if (e.cap > 0 && level[e.to] < 0) { 38 level[e.to] = level[v] + 1; 39 Q[end++] = e.to; 40 } 41 } 42 } 43 } 44 45 // ?????????? 46 int dfs(int v, int t, int f) { 47 if (v == t) return f; 48 for (int &i = iter[v]; i < G[v].size(); i++) { 49 edge &e = G[v][i]; 50 if (e.cap > 0 && level[v] < level[e.to]) { 51 int d = dfs(e.to, t, min(f, e.cap)); 52 if (d > 0) { 53 e.cap -= d; 54 G[e.to][e.rev].cap += d; 55 return d; 56 } 57 } 58 } 59 return 0; 60 } 61 62 // ?????????? 63 int Dinic(int s, int t) { 64 int flow = 0; 65 for (;;) { 66 bfs(s); 67 if (level[t] < 0) return flow; 68 memset(iter, 0, sizeof(iter)); 69 int f; 70 while ((f = dfs(s, t, INF)) > 0) { 71 flow += f; 72 } 73 } 74 } 75 76 int main() { 77 ios_base::sync_with_stdio(false); 78 79 int N, T; 80 cin >> N >> T; 81 int s = 0, t = N + T + 1; 82 83 for (int i = 1; i <= N; i++) { 84 int C; 85 cin >> C; 86 add_edge(s, i, C); 87 add_edge(i, t, 1); 88 } 89 for (int i = 1; i <= T; i++) { 90 int A, B; 91 cin >> A; 92 for (int j = 1; j <= A; j++) { 93 int x; 94 cin >> x; 95 add_edge(x, N + i, 1); 96 } 97 cin >> B; 98 for (int j = 1; j <= B; j++) { 99 int x; 100 cin >> x; 101 add_edge(N + i, x, 1); 102 } 103 } 104 105 cout << Dinic(s, t) << endl; 106 return 0; 107 }
1 #include <queue> 2 #include <cmath> 3 #include <cstdio> 4 #include <vector> 5 #include <cstring> 6 #include <iostream> 7 #include <algorithm> 8 using namespace std; 9 10 struct edge { 11 int to, cap, rev; 12 edge() {} 13 edge(int _to, int _cap, int _rev):to(_to), cap(_cap), rev(_rev) {} 14 }; 15 const int INF = 0x3f3f3f3f; 16 const int MAX_V = 250; // ????` 17 vector<edge> G[MAX_V]; // ????? 18 int level[MAX_V]; // ?????????? 19 int iter[MAX_V]; // ???????????????? 20 int Q[MAX_V]; // BFS?? 21 22 void add_edge(int from, int to, int cap) { 23 G[from].push_back(edge(to, cap, G[to].size())); 24 G[to].push_back(edge(from, 0, G[from].size() - 1)); 25 } 26 27 // ????????????????? 28 void bfs(int s) { 29 memset(level, -1, sizeof(level)); 30 int ini = 0, end = 0; 31 Q[end++] = s; 32 level[s] = 0; 33 while (end > ini) { 34 int v = Q[ini++]; 35 for (int i = 0; i < G[v].size(); i++) { 36 edge &e = G[v][i]; 37 if (e.cap > 0 && level[e.to] < 0) { 38 level[e.to] = level[v] + 1; 39 Q[end++] = e.to; 40 } 41 } 42 } 43 } 44 45 // ?????????? 46 int dfs(int v, int t, int f) { 47 if (v == t) return f; 48 for (int &i = iter[v]; i < G[v].size(); i++) { 49 edge &e = G[v][i]; 50 if (e.cap > 0 && level[v] < level[e.to]) { 51 int d = dfs(e.to, t, min(f, e.cap)); 52 if (d > 0) { 53 e.cap -= d; 54 G[e.to][e.rev].cap += d; 55 return d; 56 } 57 } 58 } 59 return 0; 60 } 61 62 // ?????????? 63 int Dinic(int s, int t) { 64 int flow = 0; 65 for (;;) { 66 bfs(s); 67 if (level[t] < 0) return flow; 68 memset(iter, 0, sizeof(iter)); 69 int f; 70 while ((f = dfs(s, t, INF)) > 0) { 71 flow += f; 72 } 73 } 74 } 75 76 int main() { 77 ios_base::sync_with_stdio(false); 78 79 int N, T; 80 cin >> N >> T; 81 int s = 0, t = N + T + 1; 82 83 for (int i = 1; i <= N; i++) { 84 int C; 85 cin >> C; 86 add_edge(s, i, C); 87 add_edge(i, t, 1); 88 } 89 for (int i = 1; i <= T; i++) { 90 int A, B; 91 cin >> A; 92 for (int j = 1; j <= A; j++) { 93 int x; 94 cin >> x; 95 add_edge(x, N + i, 1); 96 } 97 cin >> B; 98 for (int j = 1; j <= B; j++) { 99 int x; 100 cin >> x; 101 add_edge(N + i, x, 1); 102 } 103 } 104 105 cout << Dinic(s, t) << endl; 106 return 0; 107 }
Ｅ题