BestCoder9 1003 Revenge of kNN(hdu 4995) 解题报告

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 using namespace std;
 8 
 9 typedef __int64 LL;
10 const int maxn = 1e5 + 5;
11 double vi[maxn];
12 vector<int> g[maxn];
13 int N, M, K;
14 
15 struct node
16 {
17     int id, xi;
18 }point[maxn];
19 
20 int cmp(const node& a, const node& b)
21 {
22     return a.xi < b.xi;
23 }
24 
25 bool Judge(int pos, int l, int r)
26 {
27     if (r > N)
28         return true;
29     if (l <= 0)
30         return false;
31     if (point[pos].xi - point[l].xi != point[r].xi - point[pos].xi)  // 离两边距离值不同，选择距离值更小的那个点
32         return point[pos].xi - point[l].xi < point[r].xi - point[pos].xi;
33     return point[l].id < point[r].id;  // 如果离两边的距离值相同，那么选择坐标值小的那个
34 }
35 
36 void Get_KNN(int pos)   // 获得每个点的最近 k 个点是那些
37 {
38     int id = point[pos].id;
39     int l = pos - 1, r = pos + 1;
40     for (int i = 0; i < K; i++)   
41     {
42         if (Judge(pos, l, r))
43             g[id].push_back(point[l--].id);
44         else
45             g[id].push_back(point[r++].id);
46     }
47 }
48 
49 int main()
50 {
51     int T;
52     while (scanf("%d", &T) != EOF)
53     {
54         while (T--)
55         {
56             scanf("%d%d%d", &N, &M, &K);
57             for (int i = 1; i <= N; i++)
58             {
59                 scanf("%d%lf", &point[i].xi, &vi[i]);
60                 point[i].id = i;
61             }
62             sort(point+1, point+1+N, cmp);
63             for (int i = 1; i <= N; i++)
64                 Get_KNN(i);
65             int ask;
66             double ans = 0;
67             for (int i = 0; i < M; i++)
68             {
69                 scanf("%d", &ask);
70                 double sum = 0;
71                 for (int j = 0; j < g[ask].size(); j++)
72                     sum += vi[g[ask][j]];
73                 vi[ask] = sum / K;
74                 ans += vi[ask];
75             }
76             printf("%.6lf\n", ans);
77             for (int j = 1; j <= N; j++)
78                 g[j].clear();
79         }
80     }
81     return 0;
82 }