【两道水题】BZOJ 4390 [Usaco2015 dec]Max Flow & 4397 [Usaco2015 dec]Breed Counting

BZOJ 4390 [Usaco2015 dec]Max Flow

可以用树链剖分强行做，也可以用树上差分的做法。可以参考这篇blog：http://www.cnblogs.com/Created-equal/p/5111438.html

  1 #include <cstdio>
  2 #include <vector>
  3 #include <cstring>
  4 #include <algorithm>
  5 
  6 using namespace std;
  7 
  8 const size_t Max_N(50050);
  9 const size_t Max_M(100050);
 10 const int root(1);
 11 
 12 void Get_Val(int &Ret)
 13 {
 14     Ret = 0;
 15     char ch;
 16     while ((ch = getchar()), (ch > '9' || ch < '0'))
 17         ;
 18     do
 19     {
 20         (Ret *= 10) += ch -'0';
 21     }
 22     while ((ch = getchar()), (ch >= '0' && ch <= '9'));
 23 }
 24 
 25 int N;
 26 int Total;
 27 int Head[Max_N];
 28 int To[Max_M], Next[Max_M];
 29 
 30 int K;
 31 
 32 int Value[Max_N];
 33 int Father[Max_N], Deep[Max_N], Anc[Max_N][21];
 34 vector<int> Son[Max_N];
 35 
 36 int Ans;
 37 
 38 inline
 39 void Add_Edge(const int &s, const int &t)
 40 {
 41     ++Total;
 42     To[Total] = t;
 43     Next[Total] = Head[s], Head[s] = Total;
 44 }
 45 
 46 void make_tree(const int &u, const int &fa)
 47 {
 48     int v;
 49     for (int i = Head[u];i;i = Next[i])
 50     {
 51         v = To[i];
 52         if (v != fa)
 53         {
 54             Father[v] = u, Deep[v] = Deep[u] + 1;
 55             Son[u].push_back(v);
 56             make_tree(v, u);
 57         }
 58     }
 59 }
 60 
 61 void preprocess()
 62 {
 63     memset(Anc, -1, sizeof(Anc));
 64     for (int i = 1;i <= N;++i)
 65         Anc[i][0] = Father[i];
 66     for (int j = 1;(1 << j) <= N;++j)
 67         for (int i = 1;i <= N;++i)
 68             Anc[i][j] = Anc[Anc[i][j - 1]][j - 1];
 69 }
 70 
 71 int LCA(int p, int q)
 72 {
 73     if (Deep[p] < Deep[q])
 74         swap(p, q);
 75     int log;
 76     for (log = 1;(1 << log) <= Deep[p];++log); --log;
 77     for (int i = log;i >= 0;--i)
 78         if (Deep[p] - (1 << i) >= Deep[q])
 79             p = Anc[p][i];
 80     if (p == q)
 81         return p;
 82     for (int i = log;i >= 0;--i)
 83         if (Anc[p][i] != -1 && Anc[p][i] != Anc[q][i])
 84             p = Anc[p][i], q = Anc[q][i];
 85     return Father[p];
 86 }
 87 
 88 void init()
 89 {
 90     int x, y;
 91     Get_Val(N), Get_Val(K);
 92     for (int i = 1;i != N;++i)
 93     {
 94         Get_Val(x), Get_Val(y);
 95         Add_Edge(x, y), Add_Edge(y, x);
 96     }
 97     Deep[root] = 1, make_tree(root, Father[root] = -1);
 98     preprocess();
 99 }
100 
101 void Add(const int &s, const int &t, const int &w)
102 {
103     int lca = LCA(s, t);
104     Value[s] += w, Value[t] += w, Value[lca] -= w;
105     if (lca != root)
106         Value[Father[lca]] -= w;
107 }
108 
109 void DFS(const int &u)
110 {
111     for (vector<int>::size_type i = 0;i != Son[u].size();++i)
112     {
113         DFS(Son[u][i]);
114         Value[u] += Value[Son[u][i]];
115     }
116     Ans = max(Ans, Value[u]);
117 }
118 
119 void work()
120 {
121     int s, t;
122     while (K--)
123     {
124         Get_Val(s), Get_Val(t);
125         Add(s, t, 1);
126     }
127     DFS(root);
128     printf("%d", Ans);
129 }
130 
131 int main()
132 {
133     init();
134     work();
135     return 0;
136 }

BZOJ 4390

 

BZOJ 4397 [Usaco2015 dec]Breed Counting

直接用前缀和不就水过了么= =

 1 #include <cstdio>
 2 
 3 using namespace std;
 4 
 5 const size_t Max_N(100050);
 6 
 7 void Get_Val(int &Ret)
 8 {
 9     Ret = 0;
10     char ch;
11     while ((ch = getchar()), (ch > '9' || ch < '0'))
12         ;
13     do
14     {
15         (Ret *= 10) += ch - '0';
16     }
17     while ((ch = getchar()), (ch >= '0' && ch <= '9'));
18 }
19 
20 int N;
21 int Q;
22 int Prefix[4][Max_N];
23 
24 void init()
25 {
26     int Value;
27     Get_Val(N), Get_Val(Q);
28     for (int i = 1;i <= N;++i)
29     {
30         Get_Val(Value);
31         Prefix[1][i] = Prefix[1][i - 1] + (Value == 1);
32         Prefix[2][i] = Prefix[2][i - 1] + (Value == 2);
33         Prefix[3][i] = Prefix[3][i - 1] + (Value == 3);
34     }
35 }
36 
37 inline
38 int Query(const int &c, const int &a, const int &b)
39 {
40     return Prefix[c][b] - Prefix[c][a - 1];
41 }
42 
43 void work()
44 {
45     int a, b;
46     while (Q--)
47     {
48         Get_Val(a), Get_Val(b);
49         printf("%d %d %d\n", Query(1, a, b), Query(2, a, b), Query(3, a, b));
50     }
51 }
52 
53 int main()
54 {
55     init();
56     work();
57     return 0;
58 }

BZOJ 4397

 

USACO的大多数题目是真心水= =