心血来潮，用优先队列实现了这个Dijkstra，顺便练练静态链表，不过结果也蛮不错的，在acm steps里面是93msA的，在外面却是109ms，排名很靠前，好开心，不过也郁闷，竟然有人31ms过了，汗...到底是怎么样的一个牛人啊......
代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 6 const int maxn = 20000 + 100; 7 int n,m,endd; 8 typedef pair<int,int> pii; 9 struct node 10 { 11 int x,y; 12 int w; 13 }p[maxn]; 14 int nextt[maxn]; 15 int first[maxn]; 16 int d[1010]; 17 18 void dij() 19 { 20 priority_queue<pii,vector<pii>,greater<pii> > q; 21 memset(d,-1,sizeof(d)); 22 d[0] = 0; 23 q.push(make_pair(d[0],0)); 24 while(!q.empty()) 25 { 26 pii u = q.top(); 27 q.pop(); 28 int x = u.second; 29 if(u.first != d[x]) 30 continue; 31 32 for(int e = first[x];e != -1;e = nextt[e]) 33 { 34 if(d[p[e].y] == -1 || d[p[e].y] > d[x] + p[e].w) 35 { 36 d[p[e].y] = d[x] + p[e].w; 37 q.push(make_pair(d[p[e].y],p[e].y)); 38 } 39 } 40 } 41 printf("%d\n",d[endd]); 42 } 43 int main() 44 { 45 while(scanf("%d%d%d",&n,&m,&endd) == 3) 46 { 47 memset(first,-1,sizeof(first)); 48 for(int i = 0;i < m;i ++) 49 { 50 int a,b,w; 51 scanf("%d%d%d",&a,&b,&w); 52 p[i].x = a; 53 p[i].y = b; 54 p[i].w = w; 55 nextt[i] = first[a]; 56 first[a] = i; 57 } 58 int aa; 59 scanf("%d",&aa); 60 int t; 61 for(int i = 0;i < aa;i ++) 62 { 63 scanf("%d",&t); 64 p[i + m].x = 0; 65 p[i + m].y = t; 66 p[i + m].w = 0; 67 nextt[i + m] = first[0]; 68 first[0] = i + m; 69 } 70 dij(); 71 } 72 73 return 0; 74 }
