Poj 2253 Frogger (Dijkstra shortest path)

Question link: http://poj.org/problem? Id = 2253 Frogger

Time limit:1000 ms		Memory limit:65536 K
Total submissions:25773		Accepted:8374

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 login Ming 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 in 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

20 03 4317 419 418 50

Sample output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

Source

There are two frogs and several stones, one of which wants to visit the other. Now we know the coordinates of these things, the coordinates of the two frogs are the first and the second, respectively, and the frog can jump with any stone, and there are several paths between the two frogs, ask the maximum edges on all the paths, and then find the shortest path through the maximum edges. There are many paths from the start point to the end point, and each side of each path has a maximum value. Find the minimum values among these maximum values. That is, the updated edge must maintain the maximum edge. Note: I don't know much about it here, so use % in the final output. 3lf is the dead wa, and changed to %. 3f makes AC easy. In fact, it can be done on poj. If it cannot be changed, it can also be done. I 've been wa in the topic and it's helpless ~ Read the code ~

 1 #include <stdio.h> 2 #include <math.h> 3 const int INF=1010101010; 4 double map[1010][1010],node[1010],Min; 5 int n,vis[1010]; 6  7 double Max(double a,double b) 8 { 9     return a>b?a:b;10 }11 12 void dijkstra()13 {14     int i,j,k,m;15     for (i=0; i<n; i++)16     {17         node[i]=map[0][i];18         vis[i]=0;19     }20     vis[0]=1;21     for (k=0; k<n; k++)22     {23         Min=INF;24         m=-1;25         for (i=1; i<n; i++)26             if (!vis[i])27             {28                 if (Min>node[i])29                 {30                     Min=node[i];31                     m=i;32                 }33             }34         if (m==-1)35             break;36         vis[m]=1;37         //tm=m;38         for (i=0; i<n; i++)39         {40             if (!vis[i]&&Max(node[m],map[m][i])<node[i])41                 node[i]=Max(node[m],map[m][i]);42 43         }44     }45 }46 47 int main ()48 {49     double a[210],b[210];50     int count=0,i,j;51     while (scanf("%d",&n),n)52     {53         for (i=0; i<n; i++)54         {55             scanf("%lf%lf",&a[i],&b[i]);56         }57         for (i=0; i<n; i++)58         {59             for (j=0; j<n; j++)60             {61                 map[i][j]=map[j][i]=(a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]);62 63             }64         }65         dijkstra();66         printf("Scenario #%d\n", ++count);67         printf("Frog Distance = %.3f\n\n", sqrt(node[1]));68     }69     return 0;70 }