poj2253 最短路变形

题目连接：http://poj.org/problem?id=2253

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.

题目大意：

青蛙A想到B那里去，他通过跳石头的方式过去，给出A，B所在位置和石头位置的坐标，要求得出所有路径中最大距离的最小值（A要到B可以选择不同路径，每条路径跳跃若干石头，每条路径都会有一个最大跳跃距离，即该路径中最远的两块石头的距离，要求求出这些路径中最大跳跃距离的最小值）

解题思路：

最短路径变形，原来求单源最短路径算法中dis数组用来储存起点到该节点距离，现在可以用来储存到达该点的路径中最大距离的最小值，相应的松弛操作也要修改。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>

#define MAX 205
#define INF 0x3ffffff
using namespace std;

int n;
double x[MAX];
double y[MAX];
double d[MAX];
bool vis[MAX];

void init(void)
{
    memset(vis,,sizeof(vis));
    ;i<n;i++)
        d[i]=INF;
}

void dijkstra(void)
{
    d[]=;
    ;i<n;i++)
    {
        int now,m=INF;
        ;j<n;j++)
        {
            if(!vis[j]&&d[j]<m)
            {
                now=j;
                m=d[j];
            }
        }
        vis[now]=;
        ;j<n;j++)
        {
            int xx=abs(x[now]-x[j]);
            int yy=abs(y[now]-y[j]);
            double dis=sqrt(pow(xx,2.0)+pow(yy,2.0));
            d[j]=min(d[j],max(d[now],dis));
        }
    }
}

int main()
{
    ;
    while(scanf("%d",&n)!=EOF)
    {
        )
            ;
        T++;
        ;i<n;i++)
            scanf("%lf%lf",&x[i],&y[i]);
        init();
        dijkstra();
        cout<<"Scenario #"<<T<<endl;
        cout<<"Frog Distance = ";
        printf(]);
        cout<<endl;
    }
}

思考：对于类似的题目，如果是能够在节点之间转移的属性，应该都可以通过类似的方法求出来。