POJ 2253 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

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意：有n个石头，以及n个石头的坐标，青蛙1在第一块石头上，
青蛙2在第二块石头上，问青蛙1到第二块石头上， 求图中经过权值
最大值，求最小的权值（即选择的路径中，权值的最大值，都比其他
路径中权值的最大值小）。 


转化一下，就是裸地最短路了。
注意点：
1.双向路径
2.POJ的输出，用%.lf会出错

Floyd写法

  
  
Accepted
Time
32ms
Memory
688kB
Length
816
Lang
G++
Submitted
2018-01-15
 08:31:34
Shared
RemoteRunId
18042202
#include <math.h>
#include <stdio.h>
#include <string.h>
#define max(x,y) ((x)>(y)?(x):(y))
struct node{
	int x, y;
}s[205];
double e[205][205];
double Sqrt(node a, node b) {
	return sqrt((a.x-b.x)*(a.x-b.x)+ (a.y-b.y)*(a.y-b.y));
}
int main() {
	int n;
	int Case = 1;
	while(~scanf("%d",&n) && n) {
		for(int i = 1; i <= n; i++) {
			scanf("%d %d",&s[i].x, &s[i].y);
		}
		for(int i = 1; i < n; i++) {
			for(int j = i+1; j <= n; j++) {
				e[i][j] = e[j][i] = Sqrt(s[i],s[j]);
			}
		}
		//Floyd
		for(int k = 1; k <= n; k++) {
			for(int j = 1; j < n; j++) {
				for(int i = j+1; i <= n; i++) {
					if(e[j][i] > max(e[j][k] , e[k][i])) {//双向路径
						e[j][i] = e[i][j] = max(e[j][k], e[k][i]);
					}
				}
			}
		}
		printf("Scenario #%d\n",Case++);
		printf("Frog Distance = %.3f\n\n",e[1][2]);
	}
	return 0;
}
//Dijkstra

  
  
Status
Accepted
Time
16ms
Memory
692kB
Length
1077
Lang
G++
Submitted
2018-01-14
 17:42:30
Shared
RemoteRunId
18041168
#include <math.h>
#include <stdio.h>
#include <string.h>
#define inf 0x3f3f3f3f
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
struct node{
	int x, y;
}s[205];
int book[205];
double dis[205];
double e[205][205];
double Sqrt(node a, node b) {
	return sqrt((a.x-b.x)*(a.x-b.x)+ (a.y-b.y)*(a.y-b.y));
}
int main() {
	int n;
	int Case = 1;
	while(~scanf("%d",&n) && n) {
		for(int i = 1; i <= n; i++) {
			scanf("%d %d",&s[i].x, &s[i].y);
		}
		for(int i = 1; i < n; i++) {
			for(int j = i+1; j <= n; j++) {
				e[i][j] = e[j][i] = Sqrt(s[i],s[j]);
			}
		} 
		for(int i = 1; i <= n; i++) {
			dis[i] = e[1][i];
		}
		for(int i = 1; i <= n; i++) {
			book[i] = 0;
		}
		book[1] = 1;
		int p;
		double mn;
		//Dijkstra 
		for(int i = 1; i < n; i++) {
			mn = inf;
			for(int j = 1; j <= n; j++) {
				if(book[j] == 0 && dis[j] < mn) {
					mn = dis[j];
					p = j;
				}
			}
			book[p] = 1;
			for(int k = 1; k <= n; k++) {
				dis[k] = min(dis[k], max(mn, e[p][k]));//松弛
			}
		}
		printf("Scenario #%d\n",Case++);
		printf("Frog Distance = %.3f\n\n",dis[2]);
	}
	return 0;
}

Bellman-Ford
#include <math.h>
#include <stdio.h>
#include <string.h>
#define inf 0x3f3f3f3f
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
struct node{
	double x, y;
	int u, v;
	double w;
}s[40005];
double dis[205];
double Sqrt(node a, node b) {
	return sqrt((a.x-b.x)*(a.x-b.x)+ (a.y-b.y)*(a.y-b.y));
}
int main() {
	int n;
	int Case = 1;
	while(~scanf("%d",&n) && n) {
		memset(s, 0, sizeof(s));
		for(int i = 1; i <= n; i++) dis[i] = inf;	
		for(int i = 1; i <= n; i++) {
			scanf("%lf %lf",&s[i].x, &s[i].y);
		}
		int t = 0;
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j < i; j++) {
				s[t].u = i;
				s[t].v = j;
				s[t].w = Sqrt(s[i],s[j]);
				t++;	
			}
		}
		dis[1] = 0;
		for(int k = 1; k <= n; k++) {
			for(int i = 0; i < t; i++) {
				if(dis[s[i].v] > max(dis[s[i].u], s[i].w))//因为是双向路径 
				dis[s[i].v] = max(dis[s[i].u], s[i].w);
				if(dis[s[i].u] > max(dis[s[i].v], s[i].w))
				dis[s[i].u] = max(dis[s[i].v], s[i].w);
			}
		}
		printf("Scenario #%d\n",Case++);
		printf("Frog Distance = %.3f\n\n",dis[2]);
	}
	return 0;
}