poj2253 - Frogger（最短路变形）

题意杀...

有N个石头，青蛙1位于第1块石头上，青蛙2位于第2块石头上，现在青蛙要从石头1经过一些跳跃到达石头2。要求一条从石头1到石头2的跳跃路径，使路径中两个石头的最长距离是最小的。

只要将d[i]的意义变为从1到i的最大边即可，然后修改下更新条件就可以了，dij或者flo都可以

Dij

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 205
const int INF = 1<<30;
#define LL long long
int cas=1,T;
typedef struct Node
{
	double x,y;
}Point;
Point point[maxn];
double d[maxn];
int n;

double caldist(Point a,Point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void dijkstra()
{
	int vis[maxn];
	int v = 1;
	double min;
	for (int i = 1;i<=n;i++)
	{
		d[i]=INF;
		vis[i]=0;
	}
	d[1]=0;
	for (int i = 1;i<=n;i++)
	{
		min = INF;
		for (int j = 1;j<=n;j++)
		{
			if (!vis[j] && d[j] < min)
			{
				min = d[j];
				v=j;
			}
		}

		vis[v]=1;
		if (v==2)
			break;
		for (int j = 1;j<=n;j++)
			if (!vis[j] && d[j]>max(d[v],caldist(point[v],point[j])))
			{
				d[j] = max(d[v],caldist(point[v],point[j]));
			}
	}
}
int main()
{
	while (scanf("%d",&n) && n)
	{
		for (int i = 1;i<=n;i++)
			scanf("%lf%lf",&point[i].x,&point[i].y);
		dijkstra();
		printf("Scenario #%d\nFrog Distance = %.3lf\n\n",cas++,d[2]);
	}
	//freopen("in","r",stdin);
	//scanf("%d",&T);
	//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}

Floyd

#include<cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 205
#define LL long long
int cas=1,T;
struct point
{
	double x,y;
}p[205];

double dis[maxn][maxn];

int main()
{
	int n;
	while (scanf("%d",&n) && n)
	{
		printf("Scenario #%d\n",cas++);
		memset(p,0,sizeof(p));
		memset(dis,0,sizeof(dis));
		for (int i = 1;i<=n;i++)
			scanf("%lf%lf",&p[i].x,&p[i].y);
		for (int i = 1;i<=n;i++)
			for (int j = 1;j<=n;j++)
			{
				dis[i][j]=dis[j][i]=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x) + (p[i].y-p[j].y)*(p[i].y-p[j].y));
			}

		for (int k = 1;k<=n;k++)
			for (int i = 1;i<=n;i++)
				for (int j = 1;j<=n;j++)
				{
					if (dis[i][j] > max(dis[i][k],dis[k][j]))
						dis[i][j] = max(dis[i][k],dis[k][j]);
				}
		printf("Frog Distance = %.3lf\n\n",dis[1][2]);
	}
	//freopen("in","r",stdin);
	//scanf("%d",&T);
	//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}

题目

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.

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