Poj1502 spfa Shortest Path

1 // accepted 320 KB 16 MS 2 // There are n vertices, Edge Weight expressed with a 3 // give the lower Triangle Matrix, calculate the maximum value of 4 # include <cstdio> 5 # include <cstring> 6 # include <iostream> 7 # include <queue> 8 # include <cmath> 9 # include <algorithm> 10 using namespace STD; 11/** 12 * This is a documentation comment Block 13 * if you can't stick to it one day, think about why you're here! 14 * @ authr songt 15 */16 const int imax_n = 105; 17 const int imax_e = imax_n * imax_n; 18 const int INF = 100000000; 19 struct node 20 {21 int U, v, C; 22 node (INT u = 0, int V = 0, int C = 0): U (u), V (V), C (c) 23 {24 25} 26} p [imax_e]; 27 int head [imax_n]; 28 int next [imax_e]; 29 bool vis [imax_n]; 30 int dis [imax_n]; 31 int CNT [imax_n]; 32 int N; 33 int e = 0; 34 void Init () 35 {36 memset (Head,-1, sizeof (he AD); 37 memset (next,-1, sizeof (next); 38 e = 0; 39} 40 void addedge (int u, int V, int C) 41 {42 p [e] = node (u, v, c); 43 next [e] = head [u]; 44 head [u] = e ++; 45} 46 bool relax (int u, int V, int c) 47 {48 if (DIS [v]> dis [u] + C) 49 {50 dis [v] = dis [u] + C; 51 return true; 52} 53 return false; 54} 55 queue <int> q; 56 bool spfa (int src) 57 {58 While (! Q. empty () Q. pop (); 59 memset (VIS, 0, sizeof (VIS); 60 memset (CNT, 0, sizeof (CNT); 61 for (INT I = 1; I <= N; I ++) 62 dis [I] = inf; 63 dis [SRC] = 0; 64 Q. push (SRC); 65 vis [SRC] = 1; 66 CNT [SRC] ++; 67 while (! Q. empty () 68 {69 int pre = Q. front (); 70 Q. pop (); 71 vis [pre] = 0; 72 for (INT I = head [pre]; I + 1; I = next [I]) 73 {74 if (relax (PRE, P [I]. v, p [I]. c )&&! Vis [p [I]. v]) 75 {76 if (++ CNT [p [I]. v])> = N) return false; 77 vis [p [I]. v] = 1; 78 Q. push (P [I]. v); 79} 80} 81} 82 return true; 83} 84 char s [10]; 85 int main () 86 {87 while (scanf ("% d ", & N )! = EOF) 88 {89 Init (); 90 for (INT I = 2; I <= N; I ++) 91 {92 for (Int J = 1; j <I; j ++) 93 {94 scanf ("% s", S); 95 If (s [0] = 'X') continue; 96 int c = atoi (s); 97 // printf ("c = % d \ n", c); 98 addedge (I, j, C ); 99 addedge (J, I, C); 100} 101} 102 spfa (1); 103 int ans = 0; 104 for (INT I = 1; I <= N; I ++) 105 if (DIS [I]> ans) ans = dis [I]; 106 printf ("% d \ n", ANS); 107} 108 return 0; 109}

View code

 

Poj1502 spfa Shortest Path