Frogger(poj2253)
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 31854   Accepted: 10262

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 大意:
从A到B有多条路径,先取每条路径的最大值,然后再求出它们的最小值。
比如三条路径,1 3 4;2 5,1 2 1;最大值分别为4,5,2,最小值就是2 使用DIjkstra算法的思路,lowcost[]此时不再是最短路径,而是从起点到结点i的所有路径的最大边中的最小值,算法中维护s集合,从未更新的结点中取出最小节点,不断向外进行扩展。
扩展方法:if(max(lowcost[u],w[u][v])<lowcost[v])
      lowcost[v]=max(lowcost[u],w[u][v])<lowcost[v];

 #include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
#define Max 300+10
#define MMax 0x3f3f3f3f
struct point
{
int x,y;
};
double w[Max][Max];
double lowcost[Max];
bool vis[Max];
point p[Max];
int n;
double dis(point a,point b)
{
return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void Dijkstra(int s)
{
for(int i=;i<n;i++){
lowcost[i]=MMax; /*memset按照字节来赋值,但是int是四个字节*/
vis[i]=;
}
lowcost[]=;
while()
{
int u=-,v;
int i,j;
double Min=MMax;
for(v=;v<n;v++)
if(!vis[v]&&(u==-||lowcost[v]<lowcost[u]))
{
u=v;
}
if(u==-) break;
vis[u]=;
for(v=;v<n;v++)
if(!vis[v]&&max(lowcost[u],w[u][v])<lowcost[v])
lowcost[v]=max(lowcost[u],w[u][v]);
}
return;
}
int main()
{
int i,j,k,count=;
//freopen("in.txt","r",stdin);
while(cin>>n&&n)
{
memset(w,,sizeof());
for(i=;i<n;i++)
scanf("%d%d",&p[i].x,&p[i].y);
for(i=;i<n-;i++)
for(j=i+;j<n;j++)
w[i][j]=w[j][i]=dis(p[i],p[j]);
Dijkstra();
printf("Scenario #%d\nFrog Distance = %0.3lf\n\n",++count,lowcost[]);
}
}
 

Frogger(最短路)的更多相关文章

  1. POJ2253 Frogger —— 最短路变形

    题目链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissi ...

  2. POJ 2253 Frogger (最短路)

    Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28333   Accepted: 9208 Descript ...

  3. B - Frogger 最短路变形('最长路'求'最短路','最短路'求'最长路')

    http://poj.org/problem?id=2253 题目大意: 有一只可怜没人爱的小青蛙,打算去找他的女神青蛙姐姐,但是池塘水路不能走,所以只能通过蹦跶的形式到达目的地,问你从小青蛙到青蛙姐 ...

  4. POJ 2253 Frogger ( 最短路变形 || 最小生成树 )

    题意 : 给出二维平面上 N 个点,前两个点为起点和终点,问你从起点到终点的所有路径中拥有最短两点间距是多少. 分析 : ① 考虑最小生成树中 Kruskal 算法,在建树的过程中贪心的从最小的边一个 ...

  5. POJ 2253 Frogger -- 最短路变形

    这题的坑点在POJ输出double不能用%.lf而要用%.f...真是神坑. 题意:给出一个无向图,求节点1到2之间的最大边的边权的最小值. 算法:Dijkstra 题目每次选择权值最小的边进行延伸访 ...

  6. POJ 2253 Frogger 最短路 难度:0

    http://poj.org/problem?id=2253 #include <iostream> #include <queue> #include <cmath&g ...

  7. poj 2253 Frogger(最短路 floyd)

    题目:http://poj.org/problem?id=2253 题意:给出两只青蛙的坐标A.B,和其他的n-2个坐标,任一两个坐标点间都是双向连通的.显然从A到B存在至少一条的通路,每一条通路的元 ...

  8. POJ2253 frogger 最短路 floyd

    #include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#inc ...

  9. 最短路(Floyd_Warshall) POJ 2253 Frogger

    题目传送门 /* 最短路:Floyd算法模板题 */ #include <cstdio> #include <iostream> #include <algorithm& ...

随机推荐

  1. NSScanner-备

    注意:在扫描的时候,如果 空格是不需要扫描的,那么将会忽略空格. 如下代码:  1  NSString *string = @"my age is d 23    34.0";   ...

  2. IOS响应式编程框架ReactiveCocoa(RAC)使用示例-备

    ReactiveCocoa是响应式编程(FRP)在IOS中的一个实现框架,它的开源地址为:https://github.com/ReactiveCocoa/ReactiveCocoa# :在网上看了几 ...

  3. Gson解析复杂JSON对象

    例如以下格式JSON: 建立对应的Java对象,注意内部类要定义成静态的 public class HResult { public String total; public String recor ...

  4. Android中读取assets文件夹中的子文件夹内容

    文件结构如下:assets/info/info AssetManager am = this.getResources().getAssets(); InputStream input = null; ...

  5. hdu 1853 最小费用流好题 环的问题

    Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others) Tota ...

  6. (DP)House Robber

    题目: You are a professional robber planning to rob houses along a street. Each house has a certain am ...

  7. 【PAT L2-001】最短路计数

    给定一个无向带权网络,无负边,无重边和自环,每个顶点有一个正数权值.首先求特定原点s到终点d的最短路的个数:然后求所有最短路中顶点权值a[i]之和最大的那条,输出这条路径. 可用dijkstra算法求 ...

  8. JavaScript面向对象之类的继承

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  9. Unity 安卓下DLL热更新一(核心思想)

    大家都知道一谈起热更新的话首选是Ulua这个插件, 其实Unity可以使用dll热更新的,如果你实在不想用Lua来编写逻辑,0.0请下看Dll+AssetBundle如何实现热更新的.让你看完这个文章 ...

  10. 45 个非常有用的 Oracle 查询语句(转)

    这里我们介绍的是 40+ 个非常有用的 Oracle 查询语句,主要涵盖了日期操作,获取服务器信息,获取执行状态,计算数据库大小等等方面的查询.这些是所有 Oracle 开发者都必备的技能,所以快快收 ...