poj 2253 Dijkstra

Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 36854 Accepted: 11847

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

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

题意：有两只青蛙，分别在两个石头上，青蛙A想要到青蛙B那儿去.

他可以直接跳到B的石头上，也可以跳到其他石头上，再从其他石头跳到B那儿.

求青蛙从A到B的所有路径中最小的Frog Distance

例如，如果从A到B某条路径跳远的距离是2，5，6，4，则Frog Distance就是6

要求求出所有通路的最大距离Frog Distance，并把这些最大距离作比较，把最小的一个最大距离作为青蛙的最小跳远距离

对于数据2青蛙有两种方案

方案1：1-2则经过距离为2.000故此时Frog Distance=2.000

方案2：1-3-2 则经过距离分别是1.414 1.414 故此时Frog Distance=1.414

故所求的最小的Frog Distance=1.414

题解：使用dijkstra最短路径算法，在计算这个最短路径的时候计算上叙的最大跳远距离

注意：这里使用#define  定义一个常量比  const  定义的量要，赋值时间短一些

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fdouble path[300];int flag[300];double map[300][300];int n;struct coordinate{    int x,y;}a[300];double calc(int i,int j){    double x=a[i].x-a[j].x;    double y=a[i].y-a[j].y;    return sqrt(x*x+y*y);}int find_it(){    int p=-1;    double min=INF;    for(int i=1;i<=n;i++)        if(!flag[i]&&path[i]<min)            min=path[p=i];    return p;}double Dijkstra(){    for(int i=1;i<=n;i++){        flag[i]=0;        path[i]=INF;    }    path[1]=0;    int p=1;    while(p!=-1)    {        flag[p]=1;        for(int i=1;i<=n;i++){            double maxn=path[p]>map[p][i]?path[p]:map[p][i];            path[i]=path[i]<maxn?path[i]:maxn;        }        p=find_it();    }    return path[2];}int main(){    int cases=1;    //freopen("in.txt","r",stdin);    while(scanf("%d",&n),n)    {        for(int i=1;i<=n;i++){            scanf("%d%d",&a[i].x,&a[i].y);            for(int j=1;j<i;j++)                map[j][i]=map[i][j]=calc(i,j);            map[i][i]=0;        }        printf("Scenario #%d\nFrog Distance = %0.3lf\n\n",cases++,Dijkstra());    }    return 0;}