最短路Dijkstra变形——Frogger

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

output：

Scenario #1

Frog Distance = 5.000

 

Scenario #2

Frog Distance = 1.414

大意：每个案例中有n（n>=2)个坐标，第一个为Freddy，第二个为Fiona。Frog Distance表示两点间每条路径中最大石头间距的最小值。输出Freddy到Fiona的Frog Distance。

分析：单源最短路，Dijkstra的变形。

首先需要将坐标转换为边长存储。

for(int i=1;i<=n;i++)
   for(int j=1;j<=n;j++)
        {
            if(i==j) e[i][j]=0;
            else
            e[i][j]=e[j][i]=sqrt(pow(fabs((x[i]-x[j])*1.0),2)+pow(fabs((y[i]-y[j])*1.0),2));
        }

Dijkstra松弛判断变形

if(dis[j]>max(dis[u],e[u][j]))
   dis[j]=max(dis[u],e[u][j]);

完整代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
#define inf 0x3fffffff
using namespace std;
float e[205][205];
int main()
{
    int n,x[205],y[205],u,sum=0,book[205]={0};
    float dis[205],minn;
    while(cin>>n)
    {
        if(!n) break;
        sum++;
        memset(book,0,sizeof(book));
        for(int i=1;i<=n;i++)
        {
            cin>>x[i]>>y[i];
        }
//e（地图）录入
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
            if(i==j) e[i][j]=0;
            else
            e[i][j]=e[j][i]=sqrt(pow(fabs((x[i]-x[j])*1.0),2)+pow(fabs((y[i]-y[j])*1.0),2));
        }
//dis初始化
        for(int i=1;i<=n;i++)
            dis[i]=e[1][i];
        book[1]=1;
        for(int i=1;i<n;i++)
        {
//距离Freddy  frog distance  最近的点
            minn=inf;
            for(int j=1;j<=n;j++)
            {
                if(!book[j]&&minn>dis[j])
                {
                    minn=dis[j];
                    u=j;
                }
            }
            book[u]=1;
            for(int j=1;j<=n;j++)
            {
                if(dis[j]>max(dis[u],e[u][j]))
                    dis[j]=max(dis[u],e[u][j]);
            }
        }
        cout<<"Scenario #"<<sum<<endl;
        cout<<"Frog Distance = ";
        printf("%.3f\n\n",dis[2]);
    }
    return 0;
}