Frogger(最短路)

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=0;i<n;i++){
        lowcost[i]=MMax;        /*memset按照字节来赋值，但是int是四个字节*/
        vis[i]=0;
    }
    lowcost[0]=0;
    while(1)
    {
        int u=-1,v;
        int i,j;
        double Min=MMax;
        for(v=0;v<n;v++)
            if(!vis[v]&&(u==-1||lowcost[v]<lowcost[u]))
            {
                u=v;
            }
        if(u==-1)    break;
        vis[u]=1;
        for(v=0;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=0;
    //freopen("in.txt","r",stdin);
    while(cin>>n&&n)
    {
        memset(w,0,sizeof(0));
        for(i=0;i<n;i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        for(i=0;i<n-1;i++)
            for(j=i+1;j<n;j++)
                w[i][j]=w[j][i]=dis(p[i],p[j]);
        Dijkstra(0);
        printf("Scenario #%d
Frog Distance = %0.3lf

",++count,lowcost[1]);
    }
}