最小比例生成树（01分数规划）二分或迭代

最新推荐文章于 2022-10-09 10:09:10 发布

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最新推荐文章于 2022-10-09 10:09:10 发布

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原文链接：http://www.cnblogs.com/mypsq/p/4348182.html

本文探讨了一位国王为了连接各个村庄而修建水渠的问题。通过分析成本和距离的关系，利用最小生成树算法，采用二分法和迭代法求解最小比率。

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http://poj.org/problem?id=2728

Desert King

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20513		Accepted: 5764

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

Sample Output

1.000

题意：给出n个村庄的坐标(x,y,z)，z代表海拔高度，一个国王要修一些水渠，是这些村庄联通起来，任意连个村庄之间的距离是他们水平坐标的欧几里得距离，花费是海拔的高度差，问联通后的所有花费和比上所有距离和最小是多少；

分析：设第i条边的费用是c[i]，距离是l[i]，由此可得出公式：r=sigma(c[i]*x[i])/sigma(l[i]*x[i]);(x[i]=0或者1表示是否要取该边)

设在z(l)=sigma(c[i]*x[i])-l*sigma(l[i]*x[i]);

变形得：r=sigma(c[i]*x[i])/sigma(l[i]*x[i])=l+z(l)/sigma(l[i]*x[i]);

首先给l附上一个较大的值。

当r有比l更小值的充要条件是z(l)<0，而要求z(l)的最小值，就是求sigma(c[i]*x[i])-l*sigma(l[i]*x[i])最小生成树，边的权值是d[i]=c[i]-l*l[i];然后用求最小生成树的方法来更新z(l)的值;

每次当z(l)有小于0的值时，就说明还存在比当前l更小的比率，然后就把比l较小的l1来更新边权值，再次求z(l1)，当l越来越收敛于r时，z（l）也就趋于0；所以当z(l)=0的时候也就是比率最小的时候；

下面介绍两种方法一种是二分法（2400ms）和迭代法（255ms）

二分法：

首先去两个极端值left=0和right=100000即可，然后二分l值mid=（left+right）/2；当得到的z（mid）<0时，说明还有更优的答案此时right=mid，当z（mid）>0时，此时的mid是没有任何意义的，所以left=mid;知道right-left的值小于eps=0.00000001的精确度的时候mid就是最优答案；

迭代法：

首先给l随便附一个值a=0，然后建边求最小生成树z(a)的值，令b=z(l);加入fabs(b-a)<eps=0.000001的精确度时，b就是最优解，否则把b赋给a，然后重新建边求z的值，知道满足精确度是退出循环；

代码：

二分法：

#include"stdio.h"
#include"string.h"
#include"queue"
#include"stdlib.h"
#include"iostream"
#include"algorithm"
#include"string"
#include"iostream"
#include"map"
#include"math.h"
#define M 1005
#define eps 1e-6
#define inf 100000000
using namespace std;
double G[M][M],dis[M];
int use[M];
double prime(int n)
{
    double sum=0;
    memset(use,0,sizeof(use));
    int i,j;
    for(i=1;i<=n;i++)
        dis[i]=G[1][i];
    dis[1]=0;
    use[1]=1;
    for(j=1;j<n;j++)
    {
        double Min=inf;
        int tep=-1;
        for(i=1;i<=n;i++)
        {
            if(!use[i]&&Min>dis[i])
            {
                Min=dis[i];
                tep=i;
            }
        }
        if(tep==-1)
            break;
        use[tep]=1;
        sum+=Min;
        for(i=1;i<=n;i++)
        {
            if(!use[i]&&dis[i]>G[tep][i])
                dis[i]=G[tep][i];
        }
    }
    return sum;
}
double x[M],y[M],z[M];
double L[M][M],C[M][M];
int main()
{
    int n,i,j;
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
            scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                L[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                C[i][j]=fabs(z[i]-z[j]);
            }
        }
        double left=0;
        double right=10000000;
        double mid;
        while(right-left>eps)
        {
            mid=(left+right)/2;
            for(i=1;i<=n;i++)
                for(j=1;j<=n;j++)
                G[i][j]=C[i][j]-L[i][j]*mid;
            double ans=prime(n);
            if(ans<0)
                right=mid;
            else
                left=mid;
        }
        printf("%.3lf\n",mid);
    }
}

迭代法：

#include"stdio.h"
#include"string.h"
#include"queue"
#include"stdlib.h"
#include"iostream"
#include"algorithm"
#include"string"
#include"iostream"
#include"map"
#include"math.h"
#define M 1005
#define eps 1e-6
#define inf 100000000
using namespace std;
double G[M][M],dis[M];
bool use[M];
double x[M],y[M],z[M];
double L[M][M],C[M][M];
double prime(int n)//注意有几个与prime算法不同的地方
{
    int i,j,k,pre[M];
    double mini,b=0,c=0;
    for(i=1;i<=n;i++)
    {
        dis[i]=G[1][i];
        pre[i]=1;
        use[i]=false;
    }
    dis[1]=0;
    use[1]=true;
    for(i=1;i<n;i++)
    {
        mini=inf;
        k=-1;
        for(j=1;j<=n;j++)
        {
            if(!use[j]&&mini>dis[j])
            {
                mini=dis[j];
                k=j;
            }
        }
        if(mini==inf)break;
        b+=L[pre[k]][k];//*******************
        c+=C[pre[k]][k];//*******************
        use[k]=true;
        for(j=1;j<=n;j++)
        {
            if(!use[j]&&dis[j]>G[k][j])
            {
                dis[j]=G[k][j];
                pre[j]=k;//记录每个节点的前驱
            }
        }
    }
    return c/b;//返回比率
}
int main()
{
    int n,i,j;
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
            scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                L[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                C[i][j]=fabs(z[i]-z[j]);
            }
        }
        double left=0;
        double right;
        while(1)
        {
            for(i=1;i<=n;i++)
                for(j=1;j<=n;j++)
                G[i][j]=C[i][j]-L[i][j]*left;
            right=prime(n);
            if(fabs(right-left)<eps)break;
            left=right;
        }
        printf("%.3lf\n",right);
    }
    return 0;
}

转载于:https://www.cnblogs.com/mypsq/p/4348182.html