POJ 2253 Frogger【最短路变形——路径上最小的最大权】

链接：

http://poj.org/problem?id=2253

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/D

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21206		Accepted: 6903

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

Source

Ulm Local 1997

题意:

       给出一个无向图,求一条1~2的路径使得路径上的最大边权最小.

算法：

floyd 变形或者 Dijkstra 变形都可以

分析:

        floyd变形,将更新距离的过程改为取最大值即可.

        w[i][j] = min(w[i][j], max(w[i][k], w[k][j]))

ＰＳ：

　　　　这道题不像其它的最短路，是求总路径最小，而是不管总路径，只要保证路径上的最大边权在所有可以走的路径中最小就可以了所以要遍历每一条路径了Orz 如果用 Dijkstra 写有点像 Prim

code：

floyd：

/*
题意:给出一个无向图,求一条1~2的路径使得路径上的最大边权最小.

分析:floyd变形,将更新距离的过程改为取最大值即可.
*/
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

const double DNF = 2000;
const int maxn = 210;

double w[maxn][maxn];
int n;

struct Point{
    double x,y;
}p[maxn];

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

void floyd()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                w[i][j] = min(w[i][j], max(w[i][k], w[k][j])); //更新i——j路径上最小的最大边权
}

int main()
{
    int kcase = 0;
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }

        for(int i = 1; i <= n; i++)
        {
            w[i][i] = 0;
            for(int j = i+1; j <= n; j++)
            {
                w[i][j] = dist(p[i],p[j]);
                w[j][i] = dist(p[i],p[j]);
            }
        }

        floyd();
        printf("Scenario #%d\n", ++kcase);
        printf("Frog Distance = %.3lf\n\n", w[1][2]);
    }
}

Dijkstra：

/*
题意:给出一个无向图,求一条1~2的路径使得路径上的最大边权最小.

分析:dijkstra变形,将更新距离的过程改为取最大值即可.
*/
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

const double DNF = 2000;
const int maxn = 210;

double w[maxn][maxn];
double d[maxn];
int vis[maxn];
int n;
double ans;

struct Point{
    double x,y;
}p[maxn];

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

void Dijkstra()
{
    for(int i = 1; i <= n; i++) d[i] = DNF;
    d[1] = 0;
    memset(vis,0,sizeof(vis));

    for(int i = 1; i <= n; i++)
    {
        int x;
        double m = DNF;
        for(int y = 1; y <= n; y++) if(!vis[y] && d[y] <= m) m = d[x=y];
        vis[x] = 1;
        if(ans < d[x] && d[x]!= DNF) // ans 是这条路径上的最大权
        {
            ans = d[x];
        }
        if(x == 2) return; //走到目的地即可
        for(int y = 1; y <= n; y++) if(!vis[y])
            d[y] = min(d[y], w[x][y]); //更新未接入的点的dist
    }

}

int main()
{
    int kcase = 0;
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }

        for(int i = 1; i <= n; i++)
        {
            w[i][i] = 0;
            for(int j = i+1; j <= n; j++)
            {
                w[i][j] = dist(p[i],p[j]);
                w[j][i] = dist(p[i],p[j]);
            }
        }

        ans = 0;
        Dijkstra();
        printf("Scenario #%d\n", ++kcase);
        printf("Frog Distance = %.3lf\n\n", ans);
    }
}