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  • poj 2253 -- Frogger

    Frogger
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 24579   Accepted: 7975

    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
    

    一个最小生成树的变形题,或者最短路的变形题。我是用Kru做的。。

     1 /*======================================================================
     2  *           Author :   kevin
     3  *         Filename :   Frogger.cpp
     4  *       Creat time :   2014-07-05 15:33
     5  *      Description :
     6 ========================================================================*/
     7 #include <iostream>
     8 #include <algorithm>
     9 #include <cstdio>
    10 #include <cstring>
    11 #include <queue>
    12 #include <cmath>
    13 #define clr(a,b) memset(a,b,sizeof(a))
    14 #define M 205
    15 using namespace std;
    16 double c[M][M],x[M],y[M];
    17 int father[M];
    18 struct Node{
    19     int u,v;
    20     double len;
    21 }node[M*M];
    22 void make_set(int n)
    23 {
    24     for(int i = 0; i < n; i++){
    25         father[i] = i;
    26     }
    27 }
    28 int find_set(int x)
    29 {
    30     if(x != father[x]){
    31         father[x] = find_set(father[x]);
    32     }
    33     return father[x];
    34 }
    35 
    36 bool Union(int x,int y){
    37     int xx = find_set(x);
    38     int yy = find_set(y);
    39     if(xx == yy) return false;
    40     father[xx] = yy;
    41     if(find_set(0) == find_set(1)) return true;
    42     return false;
    43 }
    44 bool cmp(struct Node a,struct Node b)
    45 {
    46     return a.len < b.len;
    47 }
    48 double dist(int i,int j){
    49     return sqrt((x[i]-x[j]) * (x[i]-x[j]) + (y[i]-y[j]) * (y[i]-y[j]));
    50 }
    51 
    52 int main(int argc,char *argv[])
    53 {
    54     int n,cas = 1;
    55     while(scanf("%d",&n) && n){
    56         printf("Scenario #%d
    ",cas++);
    57         make_set(n);
    58         for(int i = 0; i < n; i++){
    59             scanf("%lf%lf",&x[i],&y[i]);
    60         }
    61         int cnt = 0;
    62         for(int i = 0; i < n; i++){
    63             for(int j = i+1; j < n; j++){
    64                 node[cnt].u = i; node[cnt].v = j;
    65                 node[cnt++].len = dist(i,j);
    66             }
    67         }
    68         sort(node,node+cnt,cmp);
    69         for(int i = 0; i < cnt; i++){
    70             if(Union(node[i].u,node[i].v)){
    71                 printf("Frog Distance = %.3lf
    
    ",node[i].len);
    72                 break;
    73             }
    74         }
    75     }
    76     return 0;
    77 }
    View Code
    Do one thing , and do it well !
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  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3829320.html
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