hdu 4081(次小生成树)

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6407 Accepted Submission(s): 2239

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

Sample Output

65.00
70.00

题意：秦朝有n个城市，需要修建一些道路让任意两个城市（每个城市都有人口）都可以联通，徐福可以用用法力修一条路，这条路不需要劳动力，秦始皇希望其他的道路总长度

B尽量短，还希望连接的两个城市人口之和A尽量大，找到最大的A/B。

题解：这题的做法是去枚举每条边，如果枚举的这条边在最小生成树上，那么结果为 person[i]+person[j]/（MST-graph[i][j]）

如果没在最小生成树上，那么加进去之后我们要删掉一条边，我们肯定是要i, j所在的环里面最大的边,这里就要用到次小生成树里面的path数组了. 结果为person[i]+person[j]/（MST-path[i][j]）因为i - j不需要花费,所以不必再加上去.

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;
typedef double type;
const int N = 1005;
const double INF = 99999999;
struct Point
{
    int x,y;
} p[N];
double graph[N][N];
int person[N];
int n,m;
int pre[N];
type path[N][N],low[N]; ///path[i][j]用于记录i到j路径上的权值最大的边
bool vis[N],used[N][N];
type prim(int pos,int n){
    memset(used,false,sizeof(used));
    memset(vis,false,sizeof(vis));
    memset(path,0,sizeof(path));
    vis[pos]=true;
    type cost = 0;
    for(int i=1;i<=n;i++){
        low[i]= graph[pos][i];
        pre[i]=1;
    }
    low[pos]=1;
    for(int i=1;i<n;i++){
        type Min = INF;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&low[j]<Min){
                pos = j;
                Min = low[j];
            }
        }
        used[pre[pos]][pos] = used[pos][pre[pos]] = true;
        cost+= Min;
        vis[pos] = true;
        for(int j=1;j<=n;j++){
            if(vis[j]&&j!=pos){
                path[pos][j] = path[j][pos] = max(low[pos],path[j][pre[pos]]);
            }
            if(!vis[j]&&low[j]>graph[pos][j]){
                low[j]=graph[pos][j];
                pre[j] = pos;
            }
        }
    }
    return cost;
}
double dis(Point a,Point b)
{
    return sqrt(1.0*((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));
}
void init()
{
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(i==j) graph[i][j] = 0;
            else graph[i][j] = INF;
        }
    }
}

int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        init();
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d",&p[i].x,&p[i].y,&person[i]);
        }
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                graph[i][j] = graph[j][i] = dis(p[i],p[j]);
            }
        }
        double MST = prim(1,n);
        double Max = -1;
        for(int i=1;i<=n;i++){ ///枚举所有的边
            for(int j=1;j<=n;j++){
                if(i!=j){
                    if(used[i][j]){ ///如果枚举的边属于最小生成树,那么结果为 A/(MST-此边)
                        Max = max(Max,(person[i]+person[j])/(MST-graph[i][j]));
                    }else{      ///如果枚举的边不属于最小生成树,那么必定要删掉最小生成树中的一条边,删掉的肯定就是i-j之间最长的那条
                        Max = max(Max,(person[i]+person[j])/(MST-path[i][j]));
                    }
                }

            }
        }
        printf("%.2lf
",Max);
    }
    return 0;
}