POJ 2253 (floyd算法)

Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 50402   Accepted: 16002 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：求第一个点到第二个点的所有通路上的最大边。

解题思路：开始用的最小生成树，结果WA了，后来仔细读了题才发现是求所有通路上的最大边，那么就好办了。floyd稍微变一下形完美解决。注意在C11的标准里面是没有%lf的，输出要用%f。附上代码

#include
#include
#include
#include
#include
using namespace std;
#define maxn 205
#define inf 0x3f3f3f3f

struct node
{
    int x,y;
} nodes[maxn];
double roads[maxn][maxn];
int n;

double dis(int x,int y,int a,int b)
{
    return sqrt((x-a)*(x-a)+(y-b)*(y-b));
}
void floyd()
{
    for(int k=0; k