Poj 2253 Frogger【kuskal变形】

Frogger

Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 35247

 

Accepted: 11334

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

Source

Ulm Local 1997

 

题目大意：给你n个坐标，找一条路径，从0号节点到1号节点，使其路径中的最长路，在其他路径方案中，值最小。

思路：

1、首先我们将两两点之间距离求出来。

2、然后我们对这些边都进行排序，从小到大的排序。

3、然后我们对于从边权值小的边开始构建树，如果构建完这条边恰好使得0号节点和1号节点从不在一个集合变成在一个集合里边，那么这条边，就是我们要找的那条边。

4、模拟上述过程，其实就是克鲁斯卡尔算法的变形。

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;struct path{    int x,y,w;}a[1050*1050];int f[1050*1050];int x[1050];int y[1050];int n;bool cmp(path a,path b){    return a.w<b.w;}int find(int a){    int r=a;    while(f[r]!=r)    r=f[r];    int i=a;    int j;    while(i!=r)    {        j=f[i];        f[i]=r;        i=j;    }    return r;}void merge(int a,int b){    int A,B;    A=find(a);    B=find(b);    if(A!=B)    f[B]=A;}int main(){    int kase=0;    int cont=0;    while(~scanf("%d",&n))    {        if(n==0)break;        int cont=0;        for(int i=0;i<n;i++)        {            scanf("%d%d",&x[i],&y[i]);        }        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                a[cont].x=i;                a[cont].y=j;                a[cont++].w=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);            }        }        for(int i=0;i<n;i++)f[i]=i;        sort(a,a+cont,cmp);        printf("Scenario #%d\n",++kase);        for(int i=0;i<cont;i++)        {            if(find(a[i].x)!=find(a[i].y))            {                merge(a[i].x,a[i].y);            }            if(find(0)==find(1))            {                printf("Frog Distance = %.3f\n",sqrt(a[i].w));                break;            }        }        printf("\n");    }}