hdoj 4081 Qin Shi Huang's National Road System 【次小生成树变形】

Qin Shi Huang's National Road System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5200    Accepted Submission(s): 1811

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

 

少个判断WA到死。。。

题意：秦始皇有N个城市，现在他要为这N个城市修N-1条路，修路的费用是路的长度，他想尽可能减少修路的费用。一个谋士有奇特的魔法，他可以用魔法不需要任何花销的变出一条路。给出 A：无花销的道路所连接两个城市的人口总数，B：所需修路的长度（当然除去无花销的）。让你求出A / B的最大值。

用Map[ i ][ j ]存储i 到 j 的距离，MST[ i ][ j ]存储 MST中i 到 j 的最大权值，用ans存储最终结果，用MSTval表示MST的结果。

思路：首先，修路费用最小——>先求MST。在这个过程中求出MST上任意两点 i 和 j 间的最大权值MST[ i ][ j ]。最后枚举所有边<i， j>，判断更新如下 （用sum记录i、j城市人口总和）

一，该边是MST里面的点，ans = max(ans, sum / (MSTval - Map[ i ][ j ]))；

二，该边不是MST里面的点，ans = max(ans, sum / (MSTval - MST[ i ][ j ]))。

算法时间复杂度(N*N)。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAXN 1010
#define INF 1<<30
using namespace std;
double Map[MAXN][MAXN];
struct rec
{
    double x, y, p;
};
rec num[MAXN];
int N;
double dis(rec a, rec b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
void getMap()
{
    for(int i = 1; i <= N; i++)
        scanf("%lf%lf%lf", &num[i].x, &num[i].y, &num[i].p);
    for(int i = 1; i <= N; i++)
    {
        Map[i][i] = 0;
        for(int j = 1; j < i; j++)
            Map[i][j] = Map[j][i] = dis(num[i], num[j]);
    }
}
double low[MAXN];
bool vis[MAXN];
double MST[MAXN][MAXN];
bool InMST[MAXN][MAXN];
int pre[MAXN];
double prime()
{
    int next;
    double mincost = 0, Min;
    memset(InMST, false, sizeof(InMST));
    memset(MST, 0, sizeof(MST));
    for(int i = 1; i <= N; i++)
    {
        vis[i] = false;
        low[i] = Map[1][i];
        pre[i] = 1;
    }
    vis[1] = true;
    for(int i = 2; i <= N; i++)
    {
        next = -1, Min = INF;
        for(int j = 1; j <= N; j++)
        {
            if(!vis[j] && Min > low[j])
            {
                Min = low[j];
                next = j;
            }
        }
        vis[next] = true;
        int fa = pre[next];
        InMST[fa][next] = InMST[next][fa] = true;
        mincost += Min;
        for(int j = 1; j <= N; j++)
        {
            if(vis[j] && j != next)//少个判断WA到死。。。
                MST[next][j] = MST[j][next] = max(MST[fa][j], low[next]);
            if(!vis[j] && low[j] > Map[next][j])
            {
                low[j] = Map[next][j];
                pre[j] = next;
            }
        }
    }
    return mincost;
}
void solve()
{
    double ans = 0;
    double MSTval = prime();
    for(int i = 1; i <= N; i++)
    {
        for(int j = 1; j < i; j++)
        {
            //if(Map[i][j] == INF) continue;
            if(InMST[i][j])//该边在MST上
                ans = max((num[i].p + num[j].p) / (MSTval - Map[i][j]), ans);
            else//不在MST上
                ans = max((num[i].p + num[j].p) / (MSTval - MST[i][j]), ans);
        }
    }
    printf("%.2lf\n", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        getMap();
        solve();
    }
    return 0;
}