（POJ 2253）Frogger 求所有可达路径中的最大边的最小值 dijkstra || floyd 变形

Frogger
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 38822 Accepted: 12492
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的路径中使得路径上的最大边权最小.

分析：
在所有路径中求最大边权的最小值。从1~2，我们用dijkstra的思想，由于所有的点之间都是直接可达的，每次我们找到当前可达的最小的边，然后慢慢的向2靠近，那么这条路径上的最大值就是结果值。
当然，我们一样可以用floyd算法在所有的情况中找到最小值。

Floyd的思想： 借助k,来更新所有的i,j。相当于比较：从i->j ,和 i->k->j这两种情况，取较优者

dijkstra:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 210;const double INF = 2010;struct point{    double x,y;};point p[maxn];double w[maxn][maxn];double ans;bool vis[maxn];double d[maxn];int n;double dis(point aa,point bb){    return sqrt((aa.x-bb.x)*(aa.x-bb.x) + (aa.y-bb.y)*(aa.y-bb.y));}void dijkstra(){    memset(vis,false,sizeof(vis));    for(int i=1;i<=n;i++) d[i] = INF;    d[1] = 0;    for(int i=1;i<=n;i++)    {        int x;        double m = INF;        for(int j=1;j<=n;j++)        {            if(!vis[j] && d[j]< m)                m = d[x=j];        }        vis[x] = true;        if(ans < d[x]) ans = d[x];        if(x == 2) return;        for(int j=1;j<=n;j++) if(!vis[j])            d[j] = min(d[j],w[x][j]);    }}int main(){    int kase=1;    while(scanf("%d",&n)!=EOF && n)    {        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] = w[j][i] = dis(p[i],p[j]);            }        }        ans = 0;        dijkstra();        printf("Scenario #%d\n",kase++);        printf("Frog Distance = %.3lf\n\n",ans);    }    return 0;}

floyd：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 210;const double INF = 2010;struct point{    double x,y;};point p[maxn];double w[maxn][maxn];int n;double dis(point aa,point bb){    return sqrt((aa.x-bb.x)*(aa.x-bb.x) + (aa.y-bb.y)*(aa.y-bb.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]));}int main(){    int kase=1;    while(scanf("%d",&n)!=EOF && n)    {        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] = w[j][i] = dis(p[i],p[j]);            }        }        floyd();        printf("Scenario #%d\n",kase++);        printf("Frog Distance = %.3lf\n\n",w[1][2]);    }    return 0;}