ACM-ICPC North America Qualifier 2014 Human Cannonball Run

题意：在一个二维平面中 一个人走路的速度是 5m/s , 平面中有n个大炮，射程固定为50m且时间为2秒  ，问你最短时间。

解题思路：最短路

解题代码：

 1 // File Name: e.cpp
 2 // Author: darkdream
 3 // Created Time: 2015年03月14日 星期六 16时47分54秒
 4
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25
26 using namespace std;
27 double  mp[200][200];
28 struct node{
29   double x, y;
30 }dapao[200],be,en ;
31 int n;
32 double dis(double x1,double x2,double y1,double y2)
33 {
34    return sqrt((x1-y1)*(x1-y1) + (x2-y2)*(x2-y2));
35 }
36 double  mi ;
37 int main(){
38    scanf("%lf %lf",&be.x,&be.y);
39    scanf("%lf %lf",&en.x,&en.y);
40
41    mi = dis(be.x,be.y,en.x,en.y)/5;
42    scanf("%d",&n);
43
44    dapao[n+1].x = en.x ;
45    dapao[n+1].y = en.y ;
46    for(int i = 1;i <= n;i ++)
47    {
48       scanf("%lf %lf",&dapao[i].x,&dapao[i].y) ;
49    }
50    for(int i= 1;i <= n;i ++)
51    {
52       for(int j = 1;j <= n+1 ; j ++)
53       {
54          double  tmp;
55          mp[i][j] = dis(dapao[i].x,dapao[i].y,dapao[j].x,dapao[j].y);
56          tmp = 2 + fabs(50.0 - mp[i][j])/5 ;
57          mp[i][j] = min(mp[i][j]/5,tmp);
58       }
59    }
60    for(int k= 1;k <= n;k ++)
61    {
62      for(int i = 1;i <= n;i ++)
63      {
64        for(int j =1 ;j <= n + 1;j ++)
65          mp[i][j] = min(mp[i][j],mp[i][k] + mp[k][j]);
66      }
67    }
68    for(int i = 1;i <= n;i ++)
69    {
70         mi = min(mi,dis(be.x,be.y,dapao[i].x,dapao[i].y)/5+mp[i][n+1]);
71    }
72    printf("%f\n",mi);
73 return 0;
74 }