Weekly Challenges

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

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原文地址：https://www.cnblogs.com/Stomach-ache/p/4031912.html