最短路(Floyd_Warshall) POJ 2253 Frogger

 

题目传送门

 1 /*  2  最短路：Floyd算法模板题  3 */  4 #include <cstdio>  5 #include <iostream>  6 #include <algorithm>  7 #include <cmath>  8 #include <cstring>  9 #include <string> 10 #include <vector> 11 using namespace std; 12 13 const int MAXN = 200 + 10; 14 const int INF = 0x3f3f3f3f; 15 double d[MAXN][MAXN]; 16 int used[MAXN]; 17 struct NODE 18 { 19 int x, y; 20 }node[MAXN]; 21 22 double dis(int j, int i) 23 { 24 return sqrt ((node[j].x - node[i].x) * (node[j].x - node[i].x) + (node[j].y - node[i].y) * (node[j].y - node[i].y)); 25 } 26 27 void Floyd_Warshall(int n) 28 { 29 for (int k=1; k<=n; ++k) 30  { 31 for (int i=1; i<=n-1; ++i) 32  { 33 for (int j=i+1; j<=n; ++j) 34  { 35 if (d[i][k] < d[i][j] && d[k][j] < d[i][j]) 36  { 37 if (d[i][k] < d[k][j]) 38 d[i][j] = d[j][i] = d[k][j]; 39 else 40 d[i][j] = d[j][i] = d[i][k]; 41  } 42  } 43  } 44  } 45 46 printf ("Frog Distance = %.3f\n", d[1][2]); 47 } 48 49 int main(void) //POJ 2253 Frogger 50 { 51 //freopen ("D.in", "r", stdin); 52 53 int n; 54 int cnt = 0; 55 int first = 1; 56 while (~scanf ("%d", &n) && n) 57  { 58 for (int i=1; i<=n; ++i) 59  { 60 scanf ("%d%d", &node[i].x, &node[i].y); 61  } 62 for (int i=1; i<=n-1; ++i) 63  { 64 for (int j=i+1; j<=n; ++j) 65  { 66 d[i][j] = d[j][i] = dis (i, j); 67  } 68  } 69 70 printf ("Scenario #%d\n", ++cnt); 71  Floyd_Warshall (n); 72 puts (""); 73  } 74 75 return 0; 76 } 77 78 79 80 81 /* 82 Scenario #1 83 Frog Distance = 5.000 84 85 Scenario #2 86 Frog Distance = 1.414 87 */