POJ-2253 Frogger （最短路）

Frogger

http://poj.org/problem?id=2253

Time Limit: 1000MS		Memory Limit: 65536K

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

题目大意：给定n个点的坐标，求点1到点2的所有路径中，最长边最小的长度？

大致思路：可以按照Dijkstra的思想，设dis[i]表示从1到i点的所有路径中，最长边最小的长度，每次不更新最短路，只更新1~i的路径上最长边最小的长度即可

【注意】POJ中，double在G++下需要用%f输出，又被坑了好久。。。

#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN=205;

int n,m1,m2,e1,e2,t;
double g[MAXN][MAXN],dis[MAXN];//dis[i]表示从1到i点的所有路径中，最长边最小的长度
double x[MAXN],y[MAXN];
bool vis[MAXN];

inline double getDist(int i,int j) {
    return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}

double Dijkstra(int sta,int des) {
    dis[sta]=0;
    for(int i=1;i<=n;++i) {
        int index=0;
        for(int j=1;j<=n;++j) {
            if(!vis[j]&&dis[j]<dis[index]) {
                index=j;
            }
        }
        if(index==des) {
            return dis[des];
        }
        vis[index]=true;
        for(int j=1;j<=n;++j) {
            double mx=max(dis[index],g[index][j]);//1~j的路径中，所有边的最大值为1~index路径中所有边的最大值 与 index到j的距离 的较大值
            if(!vis[j]&&mx<dis[j]) {
                dis[j]=mx;
            }
        }
    }
    return dis[des];
}

int main() {
    int kase=0;
    while(scanf("%d",&n),n!=0) {
        dis[0]=10000007;
        for(int i=1;i<=n;++i) {
            vis[i]=false;
            scanf("%lf%lf",x+i,y+i);
            for(int j=1;j<i;++j) {
                g[i][j]=g[j][i]=getDist(i,j);
            }
            dis[i]=g[1][i];
        }
        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++kase,Dijkstra(1,2));
    }
    return 0;
}