POJ 2253 Frogger（最短路变形）

题意：在笛卡尔坐标系内， 有若干个点，每个点互相可直达，求第一个点到第二个点所经过的最大边的最小情况。

解题思路：dijkstra变形， d[i]表示从s点到i点中的最大边。让最大的边尽可能的小。求解d[2]即可。

注意精度：交G++ 用.3f 交C++.3lf

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

Memory: 660 KB		Time: 0 MS
Language: G++		Result: Accepted

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)
#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)

int buf[10];
inline long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

inline void writenum(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
        {
            buf[p++] = i % 10;
            i /= 10;
        }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
/**************************************************************/
#define MAX_N 205
const int INF = 0x3f3f3f3f;
typedef struct
{
    int x, y;
} node;
node a[MAX_N];
int n;
int vis[MAX_N];
double d[MAX_N];
double dist(node a, node b)
{
    return sqrt(double((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)));
}

inline void init()
{
    memset(vis, 0, sizeof(vis));
    fill(d, d + n + 1, INF);
}

void dijkstra(int s)
{
    double minn;
    d[s] = 0;
    for(int i = 1 ; i <= n - 1 ; i++)
    {
        minn = INF + 0.00;
        int v = -1;
        for(int j = 1 ; j <= n ; j++)
        {
            if(!vis[j] && d[j] <minn)
            {
                minn = d[j];
                v = j;
            }
        }
        vis[v] = 1;
        if(v == -1) break;
        for(int j =1 ; j <= n ; j++)
        {
            if(!vis[j] && d[j] > max(d[v], dist(a[v], a[j])))
            {
                d[j] = max(d[v], dist(a[v], a[j]));
            }
        }
    }


}


int main()
{
    int cas = 1;
    while(scanf("%d", &n)&&n)
    {
        init();
        for(int i = 1 ; i <= n ; i++)a
        {
            a[i].x = read();
            a[i].y = read();
        }
        dijkstra(1);
        printf("Scenario #%d\n", cas++);
        printf("Frog Distance = %.3f\n\n", d[2]);

    }
    return 0;

}