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 2476: Star travel

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ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s65536K890175Standard

In the future world, one can travel by spaceship from one star to another. Because of the long distance, the fuel consumption is not linear with distance. We should take more material within our spaceship, so the weight of total spaceship increased also. The fuel consumption F can be calculated by this formula.

F= a * d + b * d * d

Here, d is the distance between any two star, a and b are parameters.

Given the coordinates of each star, you should calculate the shortest path between two star. Here, shortest means the least fuel is used.

Input

The first line of each case is the number of stars n (2<=n<=20), and the double parameters a and b. The next n lines includes three double values that are x-axis, y-axis and z-axis coordinates. The next line is one integer m. There are two integers s and t in the next m lines, s and t is star number (from 1).

Output

For each question of each case, find out the shorest path between star s and t. Print the least fuel amount from star s to star t, rounded to 3 digit after the decimal point.

Sample Input

3 1.0 2.0-12.008 82.268 174.26050.033 1.675 108.423211.106 241.439 41.52823 22 1

Sample Output

176108.94629478.812

Problem Source: skywind

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This problem is used for contest: 101 

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Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

这个题是让我比较纠结的一个题首先我没有理清思路。刚开始我还傻傻的直接求两个点的距离再输出
结果很明显的错误了。之后才发现f是与d的平方相关的一个量，a+b>c这个三角形定理不再成立，
所以两点之间的最小花费f可能是通过其他点再到终点。。


#include<cstdio>
#include<math.h>
#include<iostream>
#include<cstring>
using namespace std;
double coordinate[25][4];
double map[25][25];
double length(double x1,double y1,double z1,double x2,double y2,double z2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2));
}
int main()
{
freopen("in.txt","r",stdin);
int n;
double a,b;
while(cin>>n>>a>>b)
{
for(int i = 1;i <= n;i ++)
{
cin>>coordinate[i][1]
>>coordinate[i][2]
>>coordinate[i][3];
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
double d=length(coordinate[i][1],coordinate[i][2],coordinate[i][3],
          coordinate[j][1],coordinate[j][2],coordinate[j][3]);
map[i][j]=a*d+b*d*d;
}
}
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(map[i][k]+map[k][j]<map[i][j])
map[i][j]=map[i][k]+map[k][j];
}
}
}
int m;
cin>>m;
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
printf("%.3f\n",map[x][y]);
}
}
return 0;
}