POJ --- 2728 【最优比率生成树 --- 01分数规划】

题目链接

相关证明在另一篇文章上

题意就不说了,两个村庄之间的距离为欧几里得距离,花费是两个村庄的海拔高度.
求sigma(cost[i])/sigma(lenth[i]).
思路:
还是二分答案,使得答案不断接近我们要的那个答案.具体看代码吧.

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<stack>#include<cstdlib>#define CLR(x) memset(x,0,sizeof(x))#define ll long long int#define PI acos(-1.0)#define db double#define mod 1000000007using namespace std;const int maxn=1e3+5;const db eps=1e-7;const db inf=1e9;int n;db mapp[maxn][maxn];db low[maxn];int vis[maxn];struct point{    int x,y,z;}s[maxn];db cal(int m,int n){    return sqrt(1.0 * (s[m].x-s[n].x)*(s[m].x-s[n].x)+1.0 * (s[m].y-s[n].y)*(s[m].y-s[n].y));}db prim(int fa,db mid){    db sum=0;    for(int i=1;i<=n;i++){        low[i] = abs(s[fa].z-s[i].z)*1.0-mid*mapp[fa][i];   //d[i]数组.    }    CLR(vis);    //cout << "BUG?" << endl;    vis[fa]=1;    for(int i=1;i<n;i++){  //找点肯定只用找n-1个点啊.        db minn=inf;        int v=-1;        for(int j=1;j<=n;j++){            if(!vis[j] && minn>low[j]){                v=j;                minn=low[j];            }            //cout << "BUG?" << " " << j << " " << n << endl;        }        if(v!=-1){            sum += minn;            vis[v]=1;            for(int j=1;j<=n;j++){                db tmp=abs(s[v].z-s[j].z)*1.0-mid*mapp[v][j];                if(!vis[j] && low[j] > tmp)                    low[j]= tmp;            }        }    }    //cout << "BUG?" << endl;    return sum;}int main(){    while(~scanf("%d",&n) && n){        for(int i=1;i<=n;i++){            scanf("%d %d %d",&s[i].x,&s[i].y,&s[i].z);        }        CLR(mapp);        for(int i=1;i<=n;i++){            for(int j=1;j<=n;j++){                mapp[i][j] =  cal(i,j);            }        }           // 不断测试,r=100是最好的.        db l=0,r=100.0,mid;    //二分的范围自己定,总之越大当然肯定能出答案,但是时间相应也会增长,所以选择一个合理的范围是很重的.        while(r-l>eps){            mid = (r+l) / 2 ;            if(prim(1,mid) >= 0 ) l=mid;            else r=mid;        }        printf("%.3f\n",r);    }}