2019.2.15 t2
考虑倒过来计算最短路径长度，设dis[u]表示在最坏情况下，点u到最近的一 个出口的最短路，则p个出口的dis值都是0，答案即为dis[0]。
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cmath> 5 #include <cctype> 6 #include <queue> 7 #include <algorithm> 8 using namespace std; 9 10 #define res register int 11 #define LL long long 12 #define inf 0x3f3f3f3f 13 inline int read() 14 { 15 int x(0),f(1); char ch; 16 while(!isdigit(ch=getchar())) if(ch=='-') f=-1; 17 while(isdigit(ch)) x=x*10+ch-'0',ch=getchar(); 18 return f*x; 19 } 20 21 inline int max(int x,int y){return x>y?x:y;} 22 inline int min(int x,int y){return x<y?x:y;} 23 const int N=1000000+10; 24 int n,m,k,d,tot; 25 int head[N],ver[N<<1],nxt[N<<1],edge[N<<1]; 26 int dis[N],vis[N]; 27 inline void add(int x,int y,int z){ 28 ver[++tot]=y; nxt[tot]=head[x]; head[x]=tot; edge[tot]=z; 29 } 30 struct node{ 31 int id; 32 LL dd; 33 bool operator<(const node &n2) const { 34 return dd>n2.dd;} 35 }; 36 priority_queue<node> q; 37 38 inline LL dij() 39 { 40 while(q.size()) 41 { 42 node now=q.top(); q.pop(); 43 int x=now.id,z=now.dd; 44 if(++vis[x]>d+1) continue; 45 if(vis[x]==d+1 || !dis[x]) 46 { 47 dis[x]=z; 48 for(res i(head[x]) ; i ; i=nxt[i]) 49 { 50 int y=ver[i]; 51 if(dis[y]==inf) 52 q.push((node){ver[i],dis[x]+(LL)edge[i]}); 53 } 54 } 55 } 56 } 57 58 int main() 59 { 60 freopen("maze.in","r",stdin); 61 freopen("maze.out","w",stdout); 62 63 memset(dis,inf,sizeof(dis)); 64 n=read(); m=read(); k=read(); d=read(); 65 for(res i=1 ; i<=m ; i++) 66 { 67 int x=read(),y=read(),z=read(); 68 add(x,y,z); add(y,x,z); 69 } 70 for(res i=1 ; i<=k ; i++) 71 { 72 int x=read(); dis[x]=0; 73 q.push((node){x,0}); 74 } 75 dij(); 76 if(dis[0]==inf) puts("-1"); 77 else cout<<dis[0]<<endl; 78 79 return 0; 80 }
