pku 2728 Desert King 最有比率生成树

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

题意：

给出n个点的坐标一级每个点对应的高度，每条边两个值，a[i][j] = fabs(h[i] - h[j])高度差表示修i->j这条路需要a[i][j]的费用，b[i][j]表示路的长度。让你修路，使得所有点都能够互通并且满足总的的费用/路的总长度最小。

思路：
最小比率生成树。按这个链接所讲的01分数规划——最有比率生成树

我们二分枚举解，求最小生成树保证使得值最小。然后得到f(l) == 0的解就是我们的结果。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll long long
#define inf 0x7f7f7f7f
#define MOD 1073741824
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 150
#define N 1007
using namespace std;
/*
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
*/

//freopen("din.txt","r",stdin);

const double eps = 1e-6;

struct node
{
    double x,y,z;
}p[N];

double dis[N];
bool vt[N];
double a[N][N];
double b[N][N];
int n;

int dblcmp(double x)
{
    if (x > eps) return 1;
    else if (x < -eps) return -1;
    else return 0;
}
void init()
{
    int i,j;
    for (i = 0; i < n; ++i)
    {
        for (j = 0; j < n; ++j)
        {
            a[i][j] = b[i][j] = (i != j)*inf;
        }
    }
}
double prim(double m)
{
    int i,j;
    double res = 0;
    for (i = 0; i <= n; ++i)
    {
        vt[i] = false;
        dis[i] = a[0][i] - m*b[0][i];
    }
    dis[0] = 0; vt[0] = true;
    for (int k = 0; k < n - 1; ++k)
    {
        double MIN = inf;
        for (i = 0; i < n; ++i)
        {
            if (!vt[i] && dis[i] < MIN)
            {
                j = i;
                MIN = dis[i];

            }
        }
        res += MIN;
        vt[j] = true;
        for (i = 0; i < n; ++i)
        {
            double t = a[j][i] - m*b[j][i];//这里有点巨坑，必须是【j,i】如果是【i,j】的话会tle
            if (!vt[i] && b[j][i] != inf && dis[i] > t)
            {
                dis[i] = t;
            }
        }
    }
    return res;
}


int main()
{
     //freopen("input.txt","r",stdin);
    int i,j;

    while (~scanf("%d",&n))
    {
        if (!n) break;
        for (i = 0; i < n; ++i)
        scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);

        init();
        double r = -1;
        for (i = 0; i < n; ++i)
        {
            for (j = 0; j < n; ++j)
            {
               if (i == j) continue;
                double cb = sqrt((p[i].x - p[j].x)*(p[i].x - p[j].x) + (p[i].y - p[j].y)*(p[i].y - p[j].y));
                double ca = iabs(p[i].z - p[j].z);
                b[i][j] = b[j][i] = cb;
                a[i][j] = a[j][i] = ca;
                r = max(r,ca);
            }
        }
        double l = 0;
        double mid = 0;
        while (dblcmp(l - r) < 0)
        {
            mid = (l + r)/2;
            if (prim(mid) >= 0)
            {
                l = mid;
            }
            else r = mid;
        }
        printf("%.3lf\n",mid);
    }
    return 0;
}

　　

 Dinkelbach迭代算法解决

每次prim()求出子问题的最优解，然后代入r,直到在某个范围内即可。。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll long long
#define inf 0x7f7f7f7f
#define MOD 1073741824
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 150
#define N 1007
using namespace std;
/*
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
*/

//freopen("din.txt","r",stdin);

const double eps = 1e-6;

struct node
{
    double x,y,z;
}p[N];

double dis[N];
bool vt[N];
double a[N][N];
double b[N][N];
int pre[N];
int n;

int dblcmp(double x)
{
    if (x > eps) return 1;
    else if (x < -eps) return -1;
    else return 0;
}
void init()
{
    int i,j;
    for (i = 0; i < n; ++i)
    {
        for (j = 0; j < n; ++j)
        {
            a[i][j] = b[i][j] = (i != j)*inf;
        }
    }
}
double prim(double m)
{
    int i,j;
    double tval = 0,tdis = 0;
    for (i = 0; i <= n; ++i)
    {
        vt[i] = false;
        dis[i] = a[0][i] - m*b[0][i];
        pre[i] = 0;
    }
    dis[0] = 0; vt[0] = true;
    pre[0] = -1;
    for (int k = 0; k < n - 1; ++k)
    {
        double MIN = inf;
        for (i = 0; i < n; ++i)
        {
            if (!vt[i] && dis[i] < MIN)
            {
                j = i;
                MIN = dis[i];

            }
        }
        tval += a[pre[j]][j];//记录总价值
        tdis += b[pre[j]][j];//记录总距离
        vt[j] = true;
        for (i = 0; i < n; ++i)
        {
            double t = a[j][i] - m*b[j][i];
            if (!vt[i] && b[j][i] != inf && dis[i] > t)
            {
                pre[i] = j;
                dis[i] = t;
            }
        }
    }
    return tval/tdis;//求的解
}


int main()
{
     //freopen("input.txt","r",stdin);
    int i,j;

    while (~scanf("%d",&n))
    {
        if (!n) break;
        for (i = 0; i < n; ++i)
        scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);

        init();

        for (i = 0; i < n; ++i)
        {
            for (j = 0; j < n; ++j)
            {
               if (i == j) continue;
                double cb = sqrt((p[i].x - p[j].x)*(p[i].x - p[j].x) + (p[i].y - p[j].y)*(p[i].y - p[j].y));
                double ca = iabs(p[i].z - p[j].z);
                b[i][j] = b[j][i] = cb;
                a[i][j] = a[j][i] = ca;
            }
        }
        double s = 0,r = 0;

        while (1)
        {
            r = prim(s);
            if (dblcmp(s - r) == 0) break;
            s = r;
        }
        printf("%.3lf\n",r);
    }
    return 0;
}