Zoj 1203 swordfish

Source: Internet
Author: User

Note segmentation fault for prim implementation

`1 # include <cstdio> 2 # include <cmath> 3 4 double dis [105] [105]; 5 Double P [105] [2]; // enter the coordinate point 6 double d [105]; // The distance from one point to another is 7 double vis [105]; // remember whether the point has been accessed 8 double Len; 9 int N; 10 11 // initialize dis [105] [105] 12 Void Init () 13 {14 int I, j; 15 for (I = 1; I <= N; I ++) 16 {17 for (j = 1; j <= N; j ++) 18 {19 dis [I] [J] = dis [J] [I] = 100000000; 20} 21 vis [I] = 0; 22} 23 Len = 0.0; 24} 25 26 // calculate dis [105] [105] 27 void makedis () 28 {29 int I, J; 30 31 for (I = 1; I <= N; I ++) 32 for (j = I; j <= N; j ++) 33 {34 dis [I] [J] = dis [J] [I] = SQRT (P [I] [0]-P [J] [0]) * (p [I] [0]-P [J] [0]) + (P [I] [1]-P [J] [1]) * (p [I] [1]-P [J] [1]); 35 36} 37} 38 39 40 // use the prim algorithm to obtain the shortest distance 41 void prim () 42 {43 int I, j, k; 44 double min; 45 for (I = 1; I <= N; I ++) 46 d [I] = dis [1] [I]; 47 vis [1] = 1; 48 49 for (I = 1; I <n; I ++) 50 {51 min = 0 xfffffff; // maximum value 52 for (j = 1; j <= N; j ++) 53 {5 4 If (d [J] <min &&! Vis [J]) 55 {56 min = d [J]; 57 K = J; 58} 59} 60 vis [k] = 1; 61 Len + = d [k]; 62 for (j = 1; j <= N; j ++) 63 {64 if (! Vis [J] & D [J]> dis [k] [J]) 65 d [J] = dis [k] [J]; 66} 67 68} 69} 70 71 // main function 72 int main () 73 {74 int step; 75 int I; 76 step = 0; 77 while (scanf ("% d", & N) // end with 0 78 {79 step ++; // record case80 81 If (step! = 1) 82 printf ("\ n"); 83 Init (); 84 for (I = 1; I <= N; I ++) 85 {86 scanf ("% lf", & P [I] [0], & P [I] [1]); 87} 88 makedis (); 89 prim (); 90 printf ("case # % d: \ n", step); 91 printf ("the minimal distance is: %. 2lf \ n ", Len); 92} 93 return 0; 94}`

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