Poj 2253 & amp; zoj 1942

Frogger
Time Limit: 1000 MS Memory Limit: 65536 K
Total Submissions: 20716 Accepted: 6741

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 login Ming 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 in 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 didistance = 5.000

Scenario #2
Frog Distance = 1.414 this is a variant floyd. In fact, it is very simple to find a recursive formula. The author quickly found the recursive formula for this question, but he was wrong about storage. I used two arrays to store the coordinates of x and y, but in fact, it was done with a struct, and the result was a small error. It was still not flexible after wa for half a day! The following code is used:

# Include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # include <cmath> using namespace std; const int maxn = 300; struct node {double x; double y ;}; node edge [maxn]; double w [maxn] [maxn]; int n; double juli (node a, node B) {return sqrt (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y); // formula} void floyd () {for (int k = 0; k <n; k ++) for (int I = 0; I <n; I ++) for (int j = 0; j <n; j ++) {w [I] [j] = min (w [I] [j], max (w [I] [K], w [k] [j]); // key recursive formula} int main () {int caseno = 0; while (scanf ("% d", & n )! = EOF) {if (n = 0) break; for (int I = 0; I <n; I ++) {scanf ("% lf ", & edge [I]. x, & edge [I]. y) ;}for (int I = 0; I <n; I ++) for (int j = 0; j <n; j ++) {w [I] [j] = juli (edge [I], edge [j]);} floyd (); printf ("Scenario # % d \ n ", ++ caseno); printf ("Frog Distance = %. 3f ", w [0] [1]); printf (" \ n ");} return 0 ;}