hdu4081 Qin Shi Huang's National Road System

题目大意就是说，n个城市，修成一棵树，其中有一条边是可以用魔法的，这条边是不需要计费的，每个城市有住人，问修完所有的边的总路程是B，那条魔法修建边链接的两个城市的人口是A，求A/B的最大值。

分析：一开始就能想到的是B小、A大。因为是修成树，所以B开始就直接是最小生成树的值了，同时记录两点路径上的最大边权。在就是枚举修建的魔法边，然后删掉两点间的最大边权值（前有做记录），这是直接算就是了，取最大的A/B。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;/**********************Point*****************************/struct Point{double x,y;Point(double x=0,double y=0):x(x),y(y){}};typedef Point Vector;Vector operator + (Vector A,Vector B){return Vector(A.x + B.x,A.y + B.y);}Vector operator - (Vector A,Vector B){//向量减法return Vector(A.x - B.x,A.y - B.y);}Vector operator * (Vector A,double p){//向量数乘return Vector(A.x * p,A.y * p);}Vector operator / (Vector A,double p){//向量除实数return Vector(A.x / p,A.y / p);}int dcmp(double x){//精度正负、0的判断if (fabs(x) < eps) return 0;return x < 0?-1:1;}bool operator < (const Point& A,const Point& B){//小于符号的重载return A.x < B.x || (A.x == B.x && A.y < B.y);}bool operator == (const Point& A,const Point& B){//点重的判断return dcmp(A.x - B.x) == 0&& dcmp(A.y - B.y) == 0;}double Dot(Vector A,Vector B){//向量的点乘return A.x * B.x + A.y * B.y;}double Length(Vector A){//向量的模return sqrt(Dot(A,A));}double Angle(Vector A,Vector B){//向量的夹角return acos(Dot(A,B) / Length(A) / Length(B));}double Cross(Vector A,Vector B){//向量的叉积return A.x * B.y - A.y * B.x;}double Area2(Point A,Point B,Point C){//三角形面积return Cross(B - A,C - A);}Vector Rotate(Vector A,double rad){//向量的旋转return Vector(A.x * cos(rad) - A.y * sin(rad),A.x * sin(rad) + A.y * cos(rad));}Vector Normal(Vector A){//法向量int L = Length(A);return Vector(-A.y / L,A.x / L);}double DistanceToLine(Point p,Point A,Point B){//p到直线AB的距离Vector v1 = B - A,v2 = p - A;return fabs(Cross(v1,v2)) / Length(v1);}double DistanceToSegment(Point p,Point A,Point B){//p到线段AB的距离if (A == B) return Length(p - A);Vector v1 = B - A, v2 = p - A,v3 = p - B;if (dcmp(Dot(v1,v2) < 0)) return Length(v2);else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);else return DistanceToLine(p,A,B);}bool SegmentProperIntersection(Point A1,Point A2,Point B1,Point B2){//线段相交double c1 = Cross(A2 - A1,B1 - A1),c2 = Cross(A2 - A1,B2 - A1);double c3 = Cross(B2 - B1,A1 - B1),c4 = Cross(B2 - B1,A2 - B1);return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;}const int N = 1010;Point p[N];int A[N];int Pre[N], Mark[N];double Path[N][N], Dist[N];double G[N][N];int n, m;double Prim() {double ret = 0.0;memset(Dist, INF,sizeof Dist);memset(Path, 0,sizeof Path);memset(Mark, 0,sizeof Mark);Dist[0] = 0;Mark[0] = true;for (int i = 1;i <= n;++i) {Dist[i] = G[0][i];Pre[i] = 0;}for (int i = 1;i < n;++i) {int u = -1;for (int j = 0;j < n;++j) {if (Mark[j]) continue;if (u == -1 || Dist[j] < Dist[u]) {u = j;}}if (u == -1) return -1;Mark[u] = true;ret += Dist[u];for (int j = 0;j < n;++j) {if (Mark[j] && j != u) Path[j][u] = Path[u][j] = max(Path[j][Pre[u]], Dist[u]);if (!Mark[j]) if (Dist[j] > G[u][j]) {Dist[j] = G[u][j];Pre[j] = u;}}}return ret;}void solve() {double ans = Prim();double answer = 0.0;for (int i = 0;i < n;++i) {for (int j = i + 1;j < n;++j) {answer = max(answer, (A[i] + A[j])*1.0 / (ans - Path[i][j]));}}printf("%.2lf\n", answer);}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t;scanf("%d",&t);while(t--) {scanf("%d",&n);for (int i = 0;i < n;++i)scanf("%lf%lf%d",&p[i].x,&p[i].y,&A[i]);m = 0;memset(G, INF, sizeof G);for (int i = 0;i < n;++i) {G[i][i] = 0;for (int j = i + 1;j < n;++j) {G[i][j] = G[j][i] = Length(p[i] - p[j]);}}solve();}return 0;}