NOIP2001T4car的旅行计划

Posted 2020-09-09

洛谷传送门

一看数据就是floyed（毕竟年代久远），然而建图不是那么好贱好建，只知道三个机场，需要判断斜边来求第4个机场坐标。

往后一些麻烦的建图。

最后floyed就好。

——代码

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 int n, s, a, b;
 8 int t, x[401][5], y[402][5], T[401];
 9 double map[401][401], ans;
10 
11 inline int square(int ax)
12 {
13     return ax * ax;
14 }
15 
16 inline int dis(int ax, int ay, int bx, int by)
17 {
18     return square(ax - bx) + square(ay - by);
19 }
20 
21 int main()
22 {
23     int i, j, k;
24     scanf("%d", &n);
25     while(n--)
26     {
27         scanf("%d %d %d %d", &s, &t, &a, &b);
28         for(i = 0; i < 4 * s; i++)
29          for(j = 0; j < 4 * s; j++)
30           map[i][j] = 1000000;
31         ans = 1000000;
32         for(i = 0; i < s; i++)
33         {
34             scanf("%d %d %d %d %d %d %d", &x[i][1], &y[i][1], &x[i][2], &y[i][2], &x[i][3], &y[i][3], &T[i]);
35             if(dis(x[i][1], y[i][1], x[i][2], y[i][2]) > dis(x[i][3], y[i][3], x[i][2], y[i][2])
36             && dis(x[i][1], y[i][1], x[i][2], y[i][2]) > dis(x[i][3], y[i][3], x[i][1], y[i][1]))
37             {
38                 x[i][4] = x[i][1] + x[i][2] - x[i][3];
39                 y[i][4] = y[i][1] + y[i][2] - y[i][3];
40             }
41             else if(dis(x[i][1], y[i][1], x[i][3], y[i][3]) > dis(x[i][3], y[i][3], x[i][2], y[i][2])
42                  && dis(x[i][1], y[i][1], x[i][3], y[i][3]) > dis(x[i][2], y[i][2], x[i][1], y[i][1]))
43                  {
44                      x[i][4] = x[i][1] + x[i][3] - x[i][2];
45                      y[i][4] = y[i][1] + y[i][3] - y[i][2];
46                  }
47             else if(dis(x[i][2], y[i][2], x[i][3], y[i][3]) > dis(x[i][3], y[i][3], x[i][1], y[i][1])
48                  && dis(x[i][2], y[i][2], x[i][3], y[i][3]) > dis(x[i][2], y[i][2], x[i][1], y[i][1]))
49                  {
50                      x[i][4] = x[i][2] + x[i][3] - x[i][1];
51                      y[i][4] = y[i][2] + y[i][3] - y[i][1];
52                   }
53             map[4 * i][4 * i + 1] = map[4 * i + 1][4 * i] = sqrt(dis(x[i][1], y[i][1], x[i][2], y[i][2])) * T[i];
54             map[4 * i][4 * i + 2] = map[4 * i + 2][4 * i] = sqrt(dis(x[i][1], y[i][1], x[i][3], y[i][3])) * T[i];
55             map[4 * i][4 * i + 3] = map[4 * i + 3][4 * i] = sqrt(dis(x[i][1], y[i][1], x[i][4], y[i][4])) * T[i];
56             map[4 * i + 1][4 * i + 2] = map[4 * i + 2][4 * i + 1] = sqrt(dis(x[i][2], y[i][2], x[i][3], y[i][3])) * T[i];
57             map[4 * i + 1][4 * i + 3] = map[4 * i + 3][4 * i + 1] = sqrt(dis(x[i][2], y[i][2], x[i][4], y[i][4])) * T[i];
58             map[4 * i + 2][4 * i + 3] = map[4 * i + 3][4 * i + 2] = sqrt(dis(x[i][3], y[i][3], x[i][4], y[i][4])) * T[i];
59             for(j = 0; j < 4 * i; j++)
60              for(k = 4 * i; k < 4 * i + 4; k++)
61               map[k][j] = map[j][k] = sqrt(dis(x[j / 4][j % 4 + 1], y[j / 4][j % 4 + 1], x[k / 4][k % 4 + 1], y[k / 4][k % 4 + 1])) * t;
62         }
63         for(i = 0; i < 4 * s; i++) map[i][i] = 0;
64         for(k = 0; k < 4 * s; k++)
65          for(i = 0; i < 4 * s; i++)
66           for(j = 0; j < 4 * s; j++)
67            map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
68         for(i = 4 * a - 4; i < 4 * a; i++)
69          for(j = 4 * b - 4; j < 4 * b; j++)
70           ans = min(ans, map[i][j]);
71         printf("%.1f\n", ans);
72     }
73     return 0;
74 }

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