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
題意：從0號點，到1號點，找一條能通過的路，使得這條路中的最大的邊，比其它所有可能的路中的邊都小。(最小生成樹的最大權)
先跑一次最小生成樹(kruskal)，然後再開一個數組記錄從起點到終點這條路中的每一條邊權，(這裏用dfs查找)，最後找出最大的那條邊即可。
代碼：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
#include <iostream>
#include <vector>
#include <cmath>
#include <cstdlib>
const int MAXM=40005;
const int MAXN=205;
int fa[MAXN];
int tol;
using namespace std;
struct Edge
{
int u,v;
double w;
}edge1[MAXM];
struct Edge1
{
int v;
double w;
};
vector<Edge1> edge2[MAXN];
struct stone
{
int x,y;
}point[MAXN];
double edge3[MAXM];
void add(int u,int v,double w)
{
edge1[tol].u=u;
edge1[tol].v=v;
edge1[tol++].w=w;
}
bool cmp(Edge a,Edge b)
{
return a.w<b.w;
}
int find (int x)
{
if(fa[x]==-1)
return x;
else return fa[x]=find(fa[x]);
}
void kruskal(int n)
{
for(int i=0;i<MAXN;i++)
{
edge2[i].clear();
}
memset(fa,-1,sizeof(fa));
sort(edge1,edge1+tol,cmp);
int cnt=0;
double ans=0;
for(int i=0;i<tol;i++)
{
int u=edge1[i].u;
int v=edge1[i].v;
double w=edge1[i].w;
int t1=find(u);
int t2=find(v);
if(t1!=t2)
{
//這裏注意建雙向邊
Edge1 tmp;
tmp.v=v;
tmp.w=w;
edge2[u].push_back(tmp);
tmp.v=u;
tmp.w=w;
edge2[v].push_back(tmp);
//edge2[tol].u=u;
//edge2[tol].v=v;
//edge2[tol++].w=w;
// tol++;
fa[t1]=t2;
cnt++;
}
if(cnt==n-1)break;
}
//printf("%f\n",ans);
/*if(cnt<n-1)
return -1;
else
return ans;*/
}
int ii=0;
bool dfs(int from,int end,int pre)//用edge3數組記錄邊
{
if(from==end)
return true;
else
{
for(int i=0;i<edge2[from].size();i++)
{
if(edge2[from][i].v!=pre)
if(dfs(edge2[from][i].v,end,from))
{
edge3[ii++]=edge2[from][i].w;
return true;
}
}
}
return false;
}
int main (void)
{
int num;
int ans=1;
while(~scanf("%d",&num))
{
tol=0;
if(num==0)
break;
for(int i=0;i<num;i++)
{
scanf("%d %d",&point[i].x,&point[i].y);
}
for(int i=0;i<num;i++)
{
for(int j=i+1;j<num;j++)
{
double le=sqrt((point[i].x-point[j].x)*
(point[i].x-point[j].x)+(point[i].y-point[j].y)*
(point[i].y-point[j].y));
add(i,j,le);
}
}
kruskal(num);
dfs(0,1,0);
double maxx=-1;
for(int i=0;i<ii;i++)
{
maxx=max(edge3[i],maxx);
}
ii=0;
printf("Scenario #%d\nFrog Distance = %.3f\n\n",ans++,maxx);
}
return 0;
}