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 }