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]));
View Code
  1 // I'm the Topcoder
  2 //C
  3 #include <stdio.h>
  4 #include <stdlib.h>
  5 #include <string.h>
  6 #include <ctype.h>
  7 #include <math.h>
  8 #include <time.h>
  9 //C++
 10 #include <iostream>
 11 #include <algorithm>
 12 #include <cstdio>
 13 #include <cstdlib>
 14 #include <cmath>
 15 #include <cstring>
 16 #include <cctype>
 17 #include <stack>
 18 #include <string>
 19 #include <list>
 20 #include <queue>
 21 #include <map>
 22 #include <vector>
 23 #include <deque>
 24 #include <set>
 25 using namespace std;
 26 
 27 //*************************OUTPUT*************************
 28 #ifdef WIN32
 29 #define INT64 "%I64d"
 30 #define UINT64 "%I64u"
 31 #else
 32 #define INT64 "%lld"
 33 #define UINT64 "%llu"
 34 #endif
 35 
 36 //**************************CONSTANT***********************
 37 #define INF 0x3f3f3f3f
 38 #define eps 1e-8
 39 #define PI acos(-1.)
 40 #define PI2 asin (1.);
 41 typedef long long LL;
 42 //typedef __int64 LL;   //codeforces
 43 typedef unsigned int ui;
 44 typedef unsigned long long ui64;
 45 #define MP make_pair
 46 typedef vector<int> VI;
 47 typedef pair<int, int> PII;
 48 #define pb push_back
 49 #define mp make_pair
 50 
 51 //***************************SENTENCE************************
 52 #define CL(a,b) memset (a, b, sizeof (a))
 53 #define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
 54 #define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))
 55 
 56 //****************************FUNCTION************************
 57 template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
 58 template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
 59 template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }
 60 
 61 // aply for the memory of the stack
 62 //#pragma comment (linker, "/STACK:1024000000,1024000000")
 63 //end
 64 
 65 const int maxn = 200+10;
 66 struct node{
 67     double x;
 68     double y;
 69 };
 70 node edge[maxn];
 71 int n;
 72 double dis[maxn][maxn];
 73 double juli(node a,node b){
 74     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 75 }
 76 
 77 
 78 void floyd(){
 79     for(int k=0;k<n;k++){
 80         for(int i=0;i<n;i++){
 81             for(int j=0;j<n;j++){
 82                 dis[i][j]=min(dis[i][j],max(dis[i][k],dis[k][j]));
 83             }
 84         }
 85     }
 86 }
 87 
 88 int main(){
 89     //int n;
 90     int kase=0;
 91     while(scanf("%d",&n)!=EOF){
 92         if(n==0 )  break;
 93         for(int i=0;i<n;i++){
 94             scanf("%lf%lf",&edge[i].x,&edge[i].y);
 95         }
 96         for(int i=0;i<n;i++){
 97             for(int j=0;j<n;j++){
 98                 dis[i][j]=juli(edge[i],edge[j]);
 99             }
100         }
101         floyd();
102         printf("Scenario #%d\n",++kase);
103         printf("Frog Distance = %.3lf\n\n",dis[0][1]);
104     }
105     return 0;
106 }