poj 2728 最优比率生成树

思想的话看这里

http://www.cnblogs.com/lotus3x/archive/2009/03/21/1418480.html

二分法：

/*the length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. 最优比率生成树poj 2728目标:min{∑costi/∑leni}逼近的思想，∑costi/∑leni<=x，即 ∑(costi-x*leni)<=0    是一个单调递减函数 即求边为costi-x*leni的 MST */#include<stdio.h>#include<string.h>#include<math.h>const double inf = 1e20;const int N = 1010;const double eps = 1e-8;struct point {double x,y,z;}p[N];int n,m;int flag[N];double D[N];double len[N][N];double cost[N][N];double map[N][N];double prime(){    int i,v,k;    double ret=0,mi;    for(i=1;i<=n;i++){        flag[i]=0;        D[i]=inf;    }D[1]=0;flag[1]=1;v=1;    for(k=1;k<n;k++){        for(i=1;i<=n;i++)if(!flag[i]){            if(map[v][i]<D[i])                D[i]=map[v][i];        }        mi=inf;        for(i=1;i<=n;i++)            if(!flag[i]&&D[i]<mi)                mi=D[v=i];            flag[v]=1;            ret+=mi;    }//printf("ret=%.2lf\n",ret);return ret;}double dis(point a,point b) {return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double solve(double mid){    int i,j;        for(i=1;i<=n;i++)for(j=i+1;j<=n;j++)map[i][j]=map[j][i]=cost[i][j]-mid*len[i][j];        return prime();}int main(){int i,j,k;while(scanf("%d",&n),n){     for(i=1;i<=n;i++) scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z); for(i=1;i<=n;i++) for(j=i+1;j<=n;j++) { len[i][j]=len[j][i]=dis(p[i],p[j]); cost[i][j]=cost[j][i]=fabs(p[i].z-p[j].z); } double a=0,b; for(i=1;i<=n;i++) map[i][i]=0.0; double l=0.0,r=100.0,mid; while(fabs(l-r)>eps) { mid=(l+r)/2;                 if(solve(mid)>eps) l=mid; else r=mid; } printf("%.3lf\n",r); }return 0;}

　迭代法：不用去管上下界，随意代入即可，此时求的是∑costi/∑leni的值，代码修改一下就好，迭代法的效率灰常高啊，上面的是迭代

#include<stdio.h>#include<string.h>#include<math.h>const double inf = 1e20;const int N = 1010;const double eps = 1e-8;struct point {double x,y,z;}p[N];int n,m;int flag[N];double D[N],len[N][N],cost[N][N],map[N][N];int pre[N];double dis(point a,point b) {return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double prime(){    int i,v,k;    double cost=0,len=0,mi;memset(flag,0,sizeof(flag));    for(i=1;i<=n;i++)  D[i]=inf;     D[1]=0;flag[1]=1;v=1;    for(k=1;k<n;k++){        for(i=1;i<=n;i++)if(!flag[i]&&map[v][i]<D[i]){D[i]=map[v][i];pre[i]=v;}mi=inf;for(i=1;i<=n;i++)if(!flag[i]&&D[i]<mi)mi=D[v=i];flag[v]=1;cost+=fabs(p[pre[v]].z-p[v].z);len+=dis(p[pre[v]],p[v]);}return cost/len;}double solve(double mid){int i,j;for(i=1;i<=n;i++)for(j=i+1;j<=n;j++)map[i][j]=map[j][i]=cost[i][j]-mid*len[i][j];        return prime();}int main(){int i,j;while(scanf("%d",&n),n){for(i=1;i<=n;i++)scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);for(i=1;i<=n;i++)for(j=i+1;j<=n;j++){len[i][j]=len[j][i]=dis(p[i],p[j]);cost[i][j]=cost[j][i]=fabs(p[i].z-p[j].z);}for(i=1;i<=n;i++)map[i][i]=0.0;double a=0,b;while(1){b=solve(a);if(fabs(a-b)<eps) break;a=b;}printf("%.3lf\n",b);}return 0;}