Frogger
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 15072 |
|
Accepted: 4991 |
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
我一直是按照找到的一个指导ACMer的ACM练习做题目的，这个题目是在最短路中给出的
确实没想到怎么按照最短路来做
找了下题解:最小生成树的概念就是，连接各点的权值是所有树中最小的
所以找使得路径中最长的一跳最短的过程就是构造最小生成树的过程
prim的做法，先把1点加入集合中，然后找距离集合最近的点，加入这条边，并记录
直到2点加入集合中，停止构造。
显然当前生成树中最长的一条边就是所求的边
我以前的prim的写法跟这个题目不合适，只好找了个题解，重写的prim
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 250
5#define e 0.0000001
6double dis[MAX][MAX];
7double x[MAX],y[MAX];
8int bian[MAX][2],b1;
9int v[MAX];
10short vis[MAX];
11int n;
12double ans;
13void init()
14
{
15
int i,j;
16 for (i=1; i<=n; i++) scanf("%lf%lf",&x[i],&y[i]);
17 for (i=1; i<=n ; i++ )
18 for (j=1; j<=n ; j++ )
19 if (i!=j)
20
{
21 dis[i][j]=sqrt(fabs(x[i]-x[j])*fabs(x[i]-x[j])+fabs(y[i]-y[j])*fabs(y[i]-y[j]));
22 dis[j][i]=dis[i][j];
23
}
24
}
25void prim()
26{
27 double dmin;
28 int i,j,k;
29 vis[1]=1;
30 v[1]=1;
31 for (i=2; i<=n; i++) vis[i]=0;
32 for (i=1; i<=n ; i++ )
33 {
34 dmin=210000000;
35 for (k=1; k<=i ; k++ )
36 for (j=1; j<=n ; j++ )
37 if ((vis[j]==0)&&(j!=v[k])&&(dis[v[k]][j]-e<dmin))
38 {
39 dmin=dis[v[k]][j];
40 v[i+1]=j;
41 bian[i][0]=v[k];
42 bian[i][1]=j;
43 }
44 if (v[i+1]==2) break;
45 vis[v[i+1]]=1;
46 }
47 b1=i;
48 dmin=0;
49 for (i=1; i<=b1 ; i++ )
50 {
51 if (dis[bian[i][0]][bian[i][1]]-e>dmin)
52 {
53 dmin=dis[bian[i][0]][bian[i][1]];
54 }
55 }
56 ans=dmin;
57}
58int main()
59{
60 int t;
61 t=0;
62 scanf("%d",&n);
63 while (n!=0)
64 {
65 init();
66 prim();
67 t++;
68 printf("Scenario #%d\n",t);
69 printf("Frog Distance = %.3lf\n\n",ans);
70 scanf("%d",&n);
71 }
72 return 0;
73}
74