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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

题目链接

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

Weekly Challenges - Week 11