hdu 4081 Qin Shi Huang's National Road System (次小生成树的变形)

题目：Qin Shi Huang's National Road System

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2049    Accepted Submission(s): 746

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

 

Recommend

lcy

题意：秦始皇修路要把所有的城市都连通，每个城市有相应的人口数，每条路有相应的修路费。

          现在可以选一条magic路，修路费变为0，以A代表magic 路两端的人口数和。B代表总路费。

          选一条路作为magic路，使A/B最大。         

分析：不能用贪心，因为A与B相互制约。

          要使A/B最大，那么B应该最小。故先求出n个点的最小生成树。再枚举

 每一条边，假设最小生成树的值是B, 而枚举的那条边长度是edge[i][j],  如果这一条边已经

是属于最小生成树上的，那么最终式子的值是A/(B-edge[i][j])。如果这一条不属于最小生成

树上的， 那么添加上这条边，就会有n条边，那么就会使得有了一个环，为了使得它还是一

个生成树，就要删掉环上的一条边。 为了让生成树尽量少，那么就要删掉除了加入的那条边

以外，权值最大的那条路径。 假设删除的那个边的权值是Max[i][j], 那么就是A/(B-Max[i][j]).

即：如果把这条边当作magic road的话，那么这条边以及连接u v 的mst的边就组成了一个环了

当前这条边的权值是最大的，要使剩下的路的花费最小，那么肯定要把u v间的最长的一条边给删

去就行了，也就是找环中的第二大边了。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;

struct node
{
    double x,y;
    double peo;
}city[1010];
int vis[1010],mark[1010][1010],pre[1010];
double maps[1010][1010],dis[1010],maxedge[1010][1010];
int n;
double sum,ans;

double cal(int i,int j)
{
    double xx=(city[i].x-city[j].x)*(city[i].x-city[j].x);
    double yy=(city[i].y-city[j].y)*(city[i].y-city[j].y);
    return sqrt(xx+yy);
}
void prim()
{
    int i,j,v;
    double minc;
    sum=0;
    memset(maxedge,0,sizeof(maxedge));
    memset(pre,0,sizeof(pre));
    //dis[1]=INF;
    for(i=2;i<=n;i++)
    {
        dis[i]=maps[1][i];
        pre[i]=1;
    }
    vis[1]=1;
    for(i=1;i<n;i++)
    {
        minc=INF;
        v=1;
        for(j=1;j<=n;j++)
        {
            if(!vis[j] && dis[j]<minc)
            {
                 minc=dis[j];
                 v=j;
            }
        }
        sum+=minc;
        mark[pre[v]][v]=mark[v][pre[v]]=1;
        vis[v]=1;
        for(j=1;j<=n;j++)
        {
            if(vis[j] && j!=v)
            {
                maxedge[j][v]=maxedge[v][j]=max(dis[v],maxedge[pre[v]][j]);
            }
            if(!vis[j] && maps[v][j]<dis[j])
            {
                dis[j]=maps[v][j];
                pre[j]=v;
            }
        }
    }
}
int main()
{
    int T,i,j;
    scanf("%d",&T);
    while(T--)
    {
        memset(mark,0,sizeof(mark));
        memset(vis,0,sizeof(vis));
        memset(dis,0,sizeof(dis));
        memset(city,0,sizeof(city));
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf",&city[i].x,&city[i].y,&city[i].peo);
            for(j=1;j<i;j++)
                maps[i][j]=maps[j][i]=cal(i,j);
        }
        prim();
        ans=-1;
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                if(!mark[i][j])
                    ans=max(ans,(city[i].peo+city[j].peo)/(sum-maxedge[i][j]));
                else
                    ans=max(ans,(city[i].peo+city[j].peo)/(sum-maps[i][j]));
            }
        }
        printf("%.2lf
",ans);
    }
    return 0;
}

感想：maxedge[j][v]=maxedge[v][j]=max(dis[v],maxedge[pre[v]][j]);

这一句开始写成maxedge[j][v]=maxedge[v][j]=max(dis[v],maps[pre[v]][j]);

wa了好多好多次。。。。T_T....