Frogger POJ - 2253

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

题意：2号石头上的青蛙要去找1号石头上的青蛙，保证青蛙在整个过程中的

单次跳跃长度尽可能小（最短路的变形）。

AC代码（选c++编译， 记得每一次输出之后有一行空行）

迪杰斯特拉：

Select Code

#include <iostream>
//#include <bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#define maxn 0x3f3f3f3f
using namespace std;
double mp[210][210], dis[210];
int n, vis[210];
struct node
{
    int x, y;
} a[210];
double jisuan(int x, int y, int xx, int yy)
{
    return sqrt(double((x-xx)*(x-xx))+double((y-yy)*(y-yy)));
}
void dij(int u)
{
    int i, j;
    for(i = 1; i<=n; i++)
        dis[i] = mp[u][i];
    dis[u] = 0, vis[u] = 1;
    for(i = 0; i<n-1; i++)
    {
        int mini = maxn, k = u;
        for(j = 1; j<=n; j++)
        {
            if(dis[j]<mini&&!vis[j])
            {
                mini = dis[j];
                k = j;
            }
        }
        vis[k] = 1;
        for(j = 1; j<=n; j++)
        {
            if(dis[j]>max(dis[k],mp[k][j])&&!vis[j])
            {
                dis[j] = max(dis[k], mp[k][j]);
            }
        }
    }
}
int main()
{
    int i, j, cnt = 0;
    while(scanf("%d",&n)!=EOF)
    {
        cnt++;
        if(n==0) break;
        for(i = 1; i<=n; i++)
        {
            for(j = 1; j<=n; j++)
            {
                if(i==j) mp[i][j] = 0;
                else mp[i][j] = maxn;
            }
        }
        for(i = 1; i<=n; i++)
            scanf("%d %d",&a[i].x, &a[i].y);
        for(i = 1; i<=n-1; i++)
        {
            for(j = i+1; j<=n; j++)
            {
                double p = jisuan(a[i].x, a[i].y, a[j].x, a[j].y);
                if(mp[i][j]>p)
                    mp[i][j] = mp[j][i] = p;
               // printf("!!!%d %d %.2lf\n",i, j, mp[i][j]);
            }
        }
        memset(vis, 0, sizeof(vis));
        memset(dis, maxn, sizeof(dis));
        dij(1);
        printf("Scenario #%d\n",cnt);
        printf("Frog Distance = %.3lf\n",dis[2]);
        printf("\n");
    }
    return 0;
}

弗洛伊德：

Select Code

#include <iostream>
//#include <bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#define maxn 0x3f3f3f3f
using namespace std;
double mp[300][310];
int n;
struct node
{
    int x, y;
} a[310];
double jisuan(int x, int y, int xx, int yy)
{
    return sqrt(double((x-xx)*(x-xx))+double((y-yy)*(y-yy)));
}
void f()
{
 int k, i, j;
 for(k = 1;k<=n;k++)
 {
  for(i = 1;i<=n;i++)
  {
   for(j = 1;j<=n;j++)
   {
    if(mp[i][j]>max(mp[i][k], mp[k][j]))
    mp[i][j] = max(mp[i][k], mp[k][j]);
   }
  }
 }
}
int main()
{
    int i, j, cnt = 0;
    while(scanf("%d",&n)!=EOF)
    {
        memset(mp, 0, sizeof(mp));
        cnt++;
        if(n==0) break;
        for(i = 1; i<=n; i++)
            scanf("%d %d",&a[i].x, &a[i].y);
        for(i = 1; i<=n-1; i++)
        {
            for(j = i+1; j<=n; j++)
            {
                double p = jisuan(a[i].x, a[i].y, a[j].x, a[j].y);
                    mp[i][j] = mp[j][i] = p;
            }
        }
         f();
        printf("Scenario #%d\n",cnt);
        printf("Frog Distance = %.3lf\n",mp[2][1]);
        printf("\n");
    }
    return 0;
}