poj 2253 (dis最短路径)

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本文介绍了一个经典的最短路径问题——Frogger游戏。两只青蛙分别位于湖中不同的石头上，目标是通过一系列跳跃从一只青蛙所在的石头到达另一只青蛙所在的石头，期间可以利用湖中的其他石头作为中转站。文章详细解释了如何使用Dijkstra算法来求解青蛙间的最小跳跃距离。

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24979		Accepted: 8114

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

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Source

Ulm Local 1997

#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
double G[210][210];
int n;
struct Point{
    double x,y;
}a[210];
float dist(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void dij(){
    double dis[210];
    int vis[210];
    for(int i=1;i<=n;i++)
        vis[i]=0;
    vis[1]=1;
    dis[1]=0.0;
    for(int i=2;i<=n;i++)
        dis[i]=G[1][i];
    for(int i=1;i<n;i++){
        double min=999999.0;
        int v;
        for(int j=2;j<=n;j++)
            if(!vis[j] && dis[j]<min){
                min=dis[j];
                v=j;
            }
        vis[v]=1;
        for(int j=1;j<=n;j++){
            double tmp=(dis[v]<G[v][j] ? G[v][j] : dis[v]);     //注意这里是最短路径变形
            dis[j]= tmp<dis[j] ? tmp : dis[j];
        }
    }
    cout<<setiosflags(ios::fixed)<<setprecision(3)<<dis[2]<<endl<<endl;
}
int main(){
    int cas=1;
    while(cin>>n&&n){
        for(int i=1;i<=n;i++)
            cin>>a[i].x>>a[i].y;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                G[i][j]=dist(a[i],a[j]);
        cout<<"Scenario #"<<cas++<<endl;
        cout<<"Frog Distance = ";
        dij();
    }
    return 0;
}