Qin Shi Huang's National Road System
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1287 Accepted Submission(s): 475
Problem
Description
During
the Warring States Period of ancient China(476 BC to 221 BC), there were seven
kingdoms in China ---- they were Qi, Chu, Yan, Han, Zhao, Wei and Qin. Ying
Zheng was the king of the kingdom Qin. Through 9 years of wars, he finally
conquered all six other kingdoms and became the first emperor of a unified
China in 221 BC. That was Qin dynasty ---- the first imperial dynasty of
China(not to be confused with the Qing Dynasty, the last dynasty of China). So
Ying Zheng named himself "Qin Shi Huang" because "Shi
Huang" means "the first emperor" in Chinese.
Qin Shi Huang undertook gigantic projects, including the first version of the
Great Wall of China, the now famous city-sized mausoleum guarded by a
life-sized Terracotta Army, and a massive national road system. There is a
story about the road system:
There were n cities in China and Qin Shi Huang wanted them all be connected by
n-1 roads, in order that he could go to every city from the capital city
Xianyang.
Although Qin Shi Huang was a tyrant, he wanted the total length of all roads to
be minimum,so that the road system may not cost too many people's life. A
daoshi (some kind of monk) named Xu Fu told Qin Shi Huang that he could build a
road by magic and that magic road would cost no money and no labor. But Xu Fu
could only build ONE magic road for Qin Shi Huang. So Qin Shi Huang had to
decide where to build the magic road. Qin Shi Huang wanted the total length of
all none magic roads to be as small as possible, but Xu Fu wanted the magic
road to benefit as many people as possible ---- So Qin Shi Huang decided that
the value of A/B (the ratio of A to B) must be the maximum, which A is the
total population of the two cites connected by the magic road, and B is the
total length of none magic roads.
Would you help Qin Shi Huang?
A city can be considered as a point, and a road can be considered as a line
segment connecting two points.
Input
The
first line contains an integer t meaning that there are t test cases(t <=
10).
For each test case:
The first line is an integer n meaning that there are n cities(2 < n <=
1000).
Then n lines follow. Each line contains three integers X, Y and P ( 0 <= X,
Y <= 1000, 0 < P < 100000). (X, Y) is the coordinate of a city and P
is the population of that city.
It is guaranteed that each city has a distinct location.
Output
For
each test case, print a line indicating the above mentioned maximum ratio A/B.
The result should be rounded to 2 digits after decimal point.
Sample
Input
2
4
1 1 20
1 2 30
200 2 80
200 1 100
3
1 1 20
1 2 30
2 2 40
Sample
Output
65.00
70.00
Source
2011 Asia
Beijing Regional Contest
【思路】
最小生成树+边交换。
题目中要求：两城市P之和为A，其他城市路径长度为B，有A/B最小。
简单的想可以求出最小生成树之后一次枚举n条边使徐福同学修路然后求一遍MST，时间为O(NMlogM)。
类比于求次小生成树，我们可以做一遍MST得到总权值tot，预处理出maxcost[][]为两点之间在MST上的最长边。枚举两点ij使徐福修maxcost代表的边（这种情况一定对应着删边后的生成树总权值最小），此时有A=P[i]+P[j]，有B=tot-maxcost[i][j]，比较得ans。时间为O(N^2)。
【代码】
#include<cstdio>
#include<cmath>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std; const int maxn = +;
struct Edge{
int v,next;
double w;
}e[maxn*maxn];
int en,front[maxn];
inline void AddEdge(int u,int v,double w) {
en++; e[en].v=v; e[en].w=w; e[en].next=front[u]; front[u]=en;
} int n,m;
int x[maxn],y[maxn],p[maxn];
double maxcost[maxn][maxn]; struct Edge_Krus{
int u,v;
double w;
bool operator<(const Edge_Krus& rhs) const{
return w<rhs.w;
}
}edges[maxn*maxn];
int f[maxn];
inline int find(int x) {
return x==f[x]? x:f[x]=find(f[x]);
}
inline double dist(int i,int j) {
return sqrt((double)(x[i]-x[j])*(x[i]-x[j])+(double)(y[i]-y[j])*(y[i]-y[j]));
}
double Kruskal()
{
for(int i=;i<=n;i++)
for(int j=i+;j<=n;j++)
edges[m++]=(Edge_Krus) {i,j,dist(i,j)};
for(int i=;i<=n;i++) f[i]=i;
sort(edges,edges+m);
int cnt=;
double res=;
for(int i=;i<m;i++) {
int x=find(edges[i].u),y=find(edges[i].v);
if(x!=y) {
f[x]=y;
res += edges[i].w;
AddEdge(edges[i].u,edges[i].v,edges[i].w);
AddEdge(edges[i].v,edges[i].u,edges[i].w);
if(++cnt==n-) break;
}
}
return res;
} vector<int> nodes;
void dfs(int u,int fa,double facost) {
for(int i=;i<nodes.size();i++) {
int x=nodes[i];
maxcost[x][u]=maxcost[u][x]=max(maxcost[x][fa],facost);
}
nodes.push_back(u);
for(int i=front[u];i>=;i=e[i].next) {
int v=e[i].v;
if(v!=fa) dfs(v,u,e[i].w);
}
} int main() {
int T;
scanf("%d",&T);
while(T--)
{
en=-; m=;
memset(front,-,sizeof(front)); scanf("%d",&n);
for(int i=;i<=n;i++) scanf("%d%d%d",&x[i],&y[i],&p[i]); double tot=Kruskal(); nodes.clear();
memset(maxcost,,sizeof(maxcost));
dfs(,-,); double ans=;
for(int i=;i<=n;i++)
for(int j=i+;j<=n;j++)
ans=max(ans,(double)(p[i]+p[j])/(tot-maxcost[i][j]));
printf("%.2lf\n",ans);
}
return ;
}