A. Qin Shi Huang's National Road System
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).
2
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
241
1 20
1 2 30
200 2 80
200 1 100
31
1 20
1 2 30
2 2 40
Sample Output
65.00
70.00
先求一次最小生成树
然后暴力枚举两点，两点之间的边假设为魔法路
然后DFS去找环上面的最长边，并且用枚举得到的边来替换他
然后计算一个A/B，维护一个最大值即可
比赛的时候确实没想到啊。。
我的代码：
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#define inf 99999999
using namespace std;
struct node
{
int v1;
int v2;
double len;
};
struct point
{
double x;
double y;
double p;
};
bool map[1005][1005];
point pt[1005];
node e[1005];
vector<int>tree[1005];
double dis[1005][1005];
double dp[1005][1005];
double ans,MAX;
int out[1005];
int n;
double dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void init()
{
int i,j;
MAX=0;
for(i=1;i<=n;i++)
{
tree[i].clear();
for(j=1;j<=n;j++)
dis[i][j]=dist(pt[i],pt[j]);
}
}
double max(double a,double b)
{
if(a>b)
return a;
else
return b;
}
double prim()
{
int i,j,vx,vy,min;
double minl,ret=0,leng;
for(i=1;i<=n-1;i++)
{
e[i].v1=1;
e[i].v2=i+1;
e[i].len=dis[1][i+1];
}
for(i=1;i<=n-1;i++)
{
min=-1;
minl=inf;
for(j=i;j<=n-1;j++)
{
if(e[j].len<minl)
{
minl=e[j].len;
min=j;
}
}
if(min==-1)
break;
tree[e[min].v1].push_back(e[min].v2);
tree[e[min].v2].push_back(e[min].v1);
ret=ret+minl;
MAX=max(MAX,minl);
swap(e[i],e[min]);
vx=e[i].v2;
for(j=i+1;j<=n-1;j++)
{
vy=e[j].v2;
leng=dis[vx][vy];
if(leng<e[j].len)
{
e[j].len=leng;
e[j].v1=vx;
}
}
}
return ret;
}
void dfs(int x,int pre)
{
out[x]=pre;
int i,v;
for(i=0;i<tree[x].size();i++)
{
v=tree[x][i];
if(v!=pre)
dfs(v,x);
}
}
int main()
{
int i,t,j;
double x,y,p,A,B;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
ans=0;
for(i=1;i<=n;i++)
{
scanf("%lf%lf%lf",&x,&y,&p);
pt[i].x=x;
pt[i].y=y;
pt[i].p=p;
}
init();
double temp=prim();
for(i=1;i<=n;i++)
for(j=0;j<tree[i].size();j++)
{
int v=tree[i][j];
A=pt[i].p+pt[v].p;
B=temp-dis[i][v];
ans=max(ans,A/B);
}
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if((pt[i].p+pt[j].p)/(temp-MAX)>ans)
{
dfs(i,0);
int k=j;
B=0,A=pt[i].p+pt[j].p;
while(out[k]&&k!=i)
{
B=max(dis[k][out[k]],B);
k=out[k];
}
ans=max(ans,A/(temp-B));
}
}
}
printf("%.2lf\n",ans);
}
return 0;
}