Dijkstra happy STL version optimized by heap

# Include <stdio. h> <br/> # include <string. h> <br/> # include <vector> <br/> # include <queue> <br/> # include <iostream> <br/> using namespace STD; </P> <p> const int maxn = 201; <br/> const int INF = 0x7fffffff/2; <br/> typedef struct {int L, V;} edge; <br/> typedef vector <edge> ve; <br/> struct Vex {<br/> int V, DIS; <br/> friend bool operator <(Vex V1, vex V2) {<br/> return v1.dis> v2.dis; <br/>}< br/>}; <B R/> ve e [maxn]; <br/> priority_queue <Vex> q; </P> <p> int Dijkstra (VE * E, int N, int St, int ed) <br/>{< br/> int dis [maxn], I; <br/> bool visited [maxn] = {0 }; <br/> for (I = 0; I <n; I ++) {<br/> dis [I] = inf; <br/>}</P> <p> Vex now, next; <br/> now. dis = dis [now. V = sT] = 0; <br/> edge et; <br/> while (! Q. Empty () Q. Pop (); <br/> q. Push (now); <br/> while (! Q. Empty () {<br/> now = Q. Top (); q. Pop (); <br/> If (! Visited [now. v]) {<br/> visited [now. v] = true; <br/> for (I = 0; I <E [now. v]. size (); I ++) {<br/> ET = E [now. v] [I]; <br/> next. V = et. v; <br/> If (! Visited [et. v] & et. L + dis [now. v] <dis [et. v]) {<br/> next. dis = dis [next. V = et. v] = et. L + dis [now. v]; <br/> q. push (next); <br/>}< br/> for (I = 0; I <N; I ++) {<br/> If (DIS [I] = inf) {<br/> printf ("-1 "); <br/>} else {<br/> printf ("% d", DIS [I]); <br/>}< br/> return dis [ed]; <br/>}</P> <p> void _ test () <br/>{< br/> edge Ed; <br/> int n, m, I, A, B, C; </P> <p> scanf ("% d", & N, & M); <br/> for (I = 0; I <N; I ++) {<br/> E [I]. clear (); <br/>}< br/> for (I = 0; I <m; I ++) {<br/> scanf ("% d", & A, & B, & ED. l); <br/> ed. V = B; E [A]. push_back (ed); <br/> ed. V = A; E [B]. push_back (ed); <br/>}< br/> scanf ("% d", & A, & B); <br/> Dijkstra (E, n, a, B); <br/>}< br/> int main () <br/>{< br/>__ test (); <br/>}< br/>/* <br/> 8 9 <br/> 0 1 5 <br/> 1 2 8 <br/> 1 3 7 <br /> 1 5 4 <br/> 4 5 6 <br/> 5 6 7 <br/> 6 7 3 <br/> 1 7 2 <br/> 1 6 1 <br/> 2 3 <br/> 3 3 <br/> 0 1 5 <br/> 0 2 1 <br/> 1 2 3 <br/> 0 1 <br/> 4 5 <br/> 0 1 1 <br/> 0 2 2 <br/> 0 3 5 <br/> 1 2 1 <br/> 1 3 2 <br/> 0 1 <br/> */