POJ 2253 floyd思想

题意：求两块石头之间的青蛙的距离，即：起点到终点所有路径中最大跳跃距离的最小值

思路：可以用floyd来求解

floyd变形为dis[i][j]=MIN(dis[i][j],MAX(dis[i][k],dis[k][j]));

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

const int maxn = 200+10;
struct node{
    double x;
    double y;
};
node edge[maxn];
int n;
double dis[maxn][maxn];
double juli(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}


void floyd(){
    for(int k=0;k<n;k++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                dis[i][j]=min(dis[i][j],max(dis[i][k],dis[k][j]));
            }
        }
    }
}

int main(){
    //int n;
    int kase=0;
    while(scanf("%d",&n)!=EOF){
        if(n==0 )  break;
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&edge[i].x,&edge[i].y);
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                dis[i][j]=juli(edge[i],edge[j]);
            }
        }
        floyd();
        printf("Scenario #%d\n",++kase);
        printf("Frog Distance = %.3lf\n\n",dis[0][1]);
    }
    return 0;
}