poj2253 最小生成树最大权

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32122		Accepted: 10351

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping. 
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps. 
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence. 
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones. 

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone. 

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2

Frog Distance = 1.414

A青蛙在第一个坐标，B青蛙在第二个坐标，A要到B青蛙那，可以从任意石头之间跳，问A青蛙最少能跳多远

#include 
#include 
#include 
#include 
#include 
using namespace std;
struct edge
{
    double x,y;
} d[205];
const double inf=1e9*1.0;
double fd(edge a,edge b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double map[202][202];
int main()
{
    int n;
    int tot=1;
    bool used[205];
    double cost[202];
    while(cin>>n)
    {
        memset(used,false,sizeof(used));
        if(n==0)
            break;
        for(int i=1; i<=n; i++)
            cin>>d[i].x>>d[i].y;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
                map[i][j]=fd(d[i],d[j]);
        }
        used[1]=true;
        double ans=-inf;
        for(int i=1; i<=n; i++)
            cost[i]=map[1][i];///集合cost
        for(int i=1; icost[j])
                {
                    minc=cost[j];
                    k=j;
                }
            }
            ans=max(ans,minc);///最大权
            used[k]=true;
            if(k==2)///找到B
                break;
            for(int j=1; j<=n; j++)///更新cost 从K到j的最小距离
            {
                if(map[k][j]