最优比率生成树 POJ 2728 迭代或者二分
别人解题报告的链接：
http://blog.sina.com.cn/s/blog_691190870101626q.html
说明一下关于精度的问题，当结果是精确到小数点后3为，你自然要把误差定为至少10^(-4),我定的是10^(-8)````这里多定点没事的···
然后对于POJ上的提交，如果是用C++提交，可以写printf("%.3lf\n",ans);
但是如果是用的G++提交，就得用printf("%.3f\n",ans);
当然ans定义的是double型的·····
补充的解题报告链接：
http://blog.csdn.net/sdj222555/article/details/7490797
1 //#define debug 2 #include <cstdio> 3 #include <cmath> 4 #include <cstring> 5 using namespace std; 6 #define N 1005 7 #define eps 1e-8 8 #define INF 1e300 9 struct point 10 { 11 int x,y,z; 12 } p[N]; 13 double dis(point a,point b) 14 { 15 double t = (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); 16 return sqrt(t); 17 } 18 struct arc 19 { 20 double len,cost; 21 } edge[N][N]; 22 double lowcost[N]; 23 int pre[N]; 24 bool vis[N]; 25 int n; 26 double prim(double d) 27 { 28 double tl =0,tc=0,ra; 29 memset(vis,0,sizeof(vis)); 30 vis[0] = 1; 31 for(int i=1; i<n; ++i) 32 { 33 lowcost[i] = edge[0][i].cost-d*edge[0][i].len; 34 pre[i] = 0; 35 } 36 for(int k=1; k<n; ++k) 37 { 38 double mi = INF; 39 int v; 40 for(int i=1; i<n; ++i) 41 { 42 if(!vis[i] && lowcost[i] < mi) 43 { 44 mi = lowcost[i]; 45 v = i; 46 } 47 } 48 vis[v] = 1; 49 tc += edge[v][pre[v]].cost; 50 tl += edge[v][pre[v]].len; 51 for(int i=1; i<n; ++i) 52 { 53 if(!vis[i] && lowcost[i] > edge[v][i].cost-d*edge[v][i].len) 54 { 55 lowcost[i] = edge[v][i].cost-d*edge[v][i].len; 56 pre[i] = v; 57 } 58 } 59 } 60 ra = tc/tl; 61 return ra; 62 } 63 int main() 64 { 65 #ifdef debug 66 freopen("in.c","r",stdin); 67 #endif 68 while(scanf("%d",&n),n) 69 { 70 for(int i=0; i<n; ++i) 71 scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z); 72 for(int i=0; i<n; ++i) 73 { 74 for(int j=i+1; j<n; ++j) 75 { 76 edge[i][j].len = edge[j][i].len = dis(p[i],p[j]); 77 edge[i][j].cost = edge[j][i].cost = fabs(p[i].z-p[j].z); 78 } 79 } 80 double r=0; 81 while(true) 82 { 83 double t = prim(r); 84 if(fabs(t-r) < eps) break; 85 r = t; 86 } 87 printf("%.3f\n",r); 88 } 89 return 0; 90 }
