Desert King
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 21361 | Accepted: 5974 |
Description
David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate
ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.
After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary
channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.
His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel
between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that
each village is at a different altitude, and different channels can‘t share a lifter. Channels can intersect safely and no three villages are on the same line.
As King David‘s prime scientist and programmer, you are asked to find out the best solution to build the channels.
Input
There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the
position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.
Output
For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.
Sample Input
4 0 0 0 0 1 1 1 1 2 1 0 3 0
Sample Output
1.000
Source
01分数规划，最优比例生成树。
用两种方法：
1.二分法
二分一个答案k，
如果有sigma(Hi)/sigma(Di)<k的，即sigma(Hi)-k*sigma(Di)<0，则k要缩小。
那么我们就求出以Hi-k*Di为边权的最小生成树，看这棵树的权值和是否<0即可。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define eps 1e-6
using namespace std;
int n,m,v[1005];
double d[1005];
struct edge
{
double h,d;
}a[1005][1005];
struct village
{
double x,y,h;
}vi[1005];
double Getdis(int x,int y)
{
return sqrt((vi[x].x-vi[y].x)*(vi[x].x-vi[y].x)+(vi[x].y-vi[y].y)*(vi[x].y-vi[y].y));
}
double prim(double k)
{
for (int i=1;i<=n;i++)
v[i]=0,d[i]=100000000.0;
d[1]=0.0;
double ans=0.0;
for (int i=1;i<=n;i++)
{
double minn=100000000.0;
int x=-1;
for (int j=1;j<=n;j++)
if (!v[j]&&d[j]<minn)
{
x=j;
minn=d[j];
}
v[x]=1;
ans+=d[x];
for (int j=1;j<=n;j++)
if (!v[j]&&d[j]>a[x][j].h-k*a[x][j].d)
d[j]=a[x][j].h-k*a[x][j].d;
}
return ans;
}
void Solve()
{
double l=0.0,r=1000.0;
while (r-l>eps)
{
double m=(l+r)/(double)2;
if (prim(m)<0) r=m;
else l=m;
}
printf("%.3f\n",l);
}
double abss(double x)
{
if (x>0) return x;
return -x;
}
int main()
{
while (scanf("%d",&n)!=EOF&&n)
{
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf",&vi[i].x,&vi[i].y,&vi[i].h);
for (int i=1;i<=n;i++)
for (int j=i+1;j<=n;j++)
{
a[i][j].h=a[j][i].h=abss(vi[i].h-vi[j].h);
a[i][j].d=a[j][i].d=Getdis(i,j);
}
Solve();
}
return 0;
}
一开始为什么一直WA呢？
我对sigma(Hi)/sigma(Di)<k这个式子的变法是k*sigma(Di)-sigma(Hi)>0，因为这里是>0所以要求最大生成树！
2.Dinkelbach
这个方法是先随便给出一个答案l，然后根据这个l求最小生成树得到一个比率是k，这个k既然要比l更优，那么我们直接
用k迭代下去就可以，就不需要漫无目的的二分下去了。可以证明k最终会收敛到一个值上，这个值就是答案。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define eps 1e-6
using namespace std;
int n,m,v[1005];
double d[1005];
double aa[1005],bb[1005];
struct edge
{
double h,d;
}a[1005][1005];
struct village
{
double x,y,h;
}vi[1005];
double p,q;
double Getdis(int x,int y)
{
return sqrt((vi[x].x-vi[y].x)*(vi[x].x-vi[y].x)+(vi[x].y-vi[y].y)*(vi[x].y-vi[y].y));
}
double prim(double k)
{
for (int i=1;i<=n;i++)
v[i]=0,d[i]=100000000.0;
d[1]=0.0;
p=q=0.0;
for (int i=1;i<=n;i++)
{
double minn=100000000.0;
int x=-1;
for (int j=1;j<=n;j++)
if (!v[j]&&d[j]<minn)
{
x=j;
minn=d[j];
}
v[x]=1;
p+=aa[x],q+=bb[x];
for (int j=1;j<=n;j++)
if (!v[j]&&d[j]>a[x][j].h-k*a[x][j].d)
d[j]=a[x][j].h-k*a[x][j].d,aa[j]=a[x][j].h,bb[j]=a[x][j].d;
}
return p/q;
}
void Solve()
{
double l=10.0;
while (1)
{
double k=prim(l);
if (fabs(k-l)<eps) break;
l=k;
}
printf("%.3f\n",l);
}
double abss(double x)
{
if (x>0) return x;
return -x;
}
int main()
{
while (scanf("%d",&n)!=EOF&&n)
{
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf",&vi[i].x,&vi[i].y,&vi[i].h);
for (int i=1;i<=n;i++)
for (int j=i+1;j<=n;j++)
{
a[i][j].h=a[j][i].h=abss(vi[i].h-vi[j].h);
a[i][j].d=a[j][i].d=Getdis(i,j);
}
Solve();
}
return 0;
}
这个方法要快很多。
但是有的题目是不适合用此法的，比如【POJ 3621】求负环这道题，很难去记录当前的比率是多少。
感悟：
1.在进行二分判断的时候，要注意如果是<0，就要找最小的看是否<0；如果是>0就要找最大的，看是否>0
2.对于Dinkelbach的证明，可以看这里。
我的大概理解是：通过当前的k值求得的新的比率要比k更优，我们一直取更优的，最终就会取到最优的了