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個指導ACMer的ACM練習做題目的，這個題目是在最短路中給出的
確實沒想到怎么按照最短路來做
找了下題解:最小生成樹的概念就是，連接各點的權值是所有樹中最小的
所以找使得路徑中最長的一跳最短的過程就是構造最小生成樹的過程
prim的做法，先把1點加入集合中，然后找距離集合最近的點，加入這條邊，并記錄
直到2點加入集合中，停止構造。
顯然當前生成樹中最長的一條邊就是所求的邊
我以前的prim的寫法跟這個題目不合適，只好找了個題解，重寫的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