Frogger
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 61596 | Accepted: 19242 |
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
0Sample Output
Scenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414Source
问题链接：POJ2253 Frogger
问题描述：青蛙Freddy和Fiona在一条河中，河中有n个石头，Freddy在1号石头上，Fiona在2号石头上，Freddy想去见Fiona，由于河水很脏，Freddy只能通过跳跃到达目的地。从石头i(x1,y1)到石头j(x2,y2)的距离为i和j的二维笛卡尔坐标距离。对于一种可行的路线，取跳跃距离最大的一步，可能有多种可行路线，最终答案是取各个可行路线中最大一步的最小值。
解题思路：并查集+贪心，求出n个点之间的相互距离，按距离从小到大排序，然后从小到大加边，直到石头1和石头2相连，那么最后添加的一条边一定是所求值
AC的C++程序：
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=205;
int pre[N];
//初始化函数 ：使各个结点是自己的父节点
void init(int n)
{
for(int i=1;i<=n;i++)
pre[i]=i;
}
int find(int x)
{
int r=x;
while(r!=pre[r])
r=pre[r];
//路径压缩
while(x!=r){
int i=pre[x];
pre[x]=r;
x=i;
}
return r;
}
struct Edge{
double l;
int u,v;
bool operator<(const Edge &a)const
{
return l<a.l;
}
}e[N*N/2];
void join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
pre[fx]=fy;
}
int main()
{
int n,cnt=0;
while(scanf("%d",&n)&&n){
double p[N][2];
for(int i=1;i<=n;i++)
scanf("%lf%lf",&p[i][0],&p[i][1]);
//计算n个点相互之间的距离
int k=0;
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++){
e[k].l=sqrt((p[i][0]-p[j][0])*(p[i][0]-p[j][0])+(p[i][1]-p[j][1])*(p[i][1]-p[j][1]));
e[k].u=i;
e[k].v=j;
k++;
}
sort(e,e+k);
init(n);
//从小到大加边
for(int i=0;i<k;i++){
join(e[i].u,e[i].v);//添加边d[i]
if(find(1)==find(2)){//添加边d[i]后1号结点和2号结点连通了，这就是答案
printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++cnt,e[i].l);
break;
}
}
}
return 0;
}