POJ 2253 Frogger解題レポート
22627 ワード
分類:図論、最短、生成ツリー
作成者:
ACShiryu
時間:2011-7-28
原題:http://poj.org/problem?id=2253
Frogger
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 13595
Accepted: 4521
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
Sample Output
参照コード:
作成者:
ACShiryu
時間:2011-7-28
原題:http://poj.org/problem?id=2253
Frogger
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 13595
Accepted: 4521
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 ， ， ， A B ， B ， ， B ， A B Frog Distance， Frog Distance A B ， ， A B 2，5，6，4， Frog Distance 6， ， 0 ， n ， 2，3 A，B ， n-2 ， （ ）， ， 。 1 5.000 2 1：1-2 2.000 Frog Distance=2.000 2：1-3-2 1.414 1.414 Frog Distance=1.414 Frog Distance=1.414 POJ1797 ， ， ， Kruskal+ ， ， Prim ， ， 0（ 0 ）， ， ， ， ， ， 1 ， 。 ，WA ， ， ， ；
参照コード:
1 #include<iostream>
2 #include<cstdlib>
3 #include<cstdio>
4 #include<cstring>
5 #include<algorithm>
6 #include<cmath>
7 using namespace std;
8 pair<int ,int> a[200]; // n
9 double lowcost[200],closet[200];//Prim ，lowcost[i] i ，closet[i] i
10 double map[200][200]; //
11 int main()
12 {
13 int n;
14 int k=1;
15 while(cin>>n,n)
16 {
17 int i,j;
18 for (i = 0 ;i < n ; i++ )
19 cin>>a[i].first>>a[i].second; // n ， 0 ， 0-1 Frog Distance
20 memset(lowcost,0,sizeof(lowcost)); //
21 for ( i = 0 ; i < n ; i ++ )
22 {
23 for ( j = 0 ; j < n ; j ++ )
24 {// ， map
25 map[i][j]=1.0*sqrt(pow(1.0*abs(a[i].first-a[j].first),2)+pow(1.0*abs(a[i].second-a[j].second),2));
26 }
27 }
28 double ans=0.0;// ， 0
29 for ( i = 0 ; i< n ; i++ )
30 {// 0 ， 0
31 lowcost[i]=map[0][i];
32 closet[i]=0;
33 }
34
35 for ( i = 0 ; i < n - 1 ; i ++ )
36 {
37 double mindis=1.0*(1<<20); // ，
38 int minone; //
39 for ( j = 0 ; j < n ; j ++ )
40 {
41 if(lowcost[j]&&mindis>lowcost[j])
42 {//j ， j ，
43 mindis=lowcost[j];
44 minone=j;
45 }
46 }
47 if(ans<mindis) //
48 ans=mindis; //
49 lowcost[minone]=0.0;//
50 if(minone==1) // 1，
51 break;
52 for ( j = 0 ; j < n ; j ++ )
53 {//
54 if(map[j][minone]<lowcost[j])
55 {// minone j j ， minone
56 lowcost[j]=map[j][minone];
57 closet[j]=minone;
58 }
59 }
60 }
61 cout<<"Scenario #"<<k<<endl;
62 printf("Frog Distance = %.3f
",ans);
63 k++;
64 }
65 return 0;
66 }