BZOJ2763 [JLOI2011]飞行路线[最短路]

显然分层图裸题，做这题纯粹是为了填坑，不解释。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #define dbg(x) cerr << #x << " = " << x <<endl
 8 using namespace std;
 9 typedef long long ll;
10 typedef double db;
11 typedef pair<int,int> pii;
12 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
13 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
14 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,1):0;}
15 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
16 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
17 template<typename T>inline T read(T&x){
18     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
19     while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
20 }
21 const int N=10000+2,M=50000+2,INF=0x3f3f3f3f;
22 struct thxorz{
23     int to,nxt,w;
24 }G[M*2*22];
25 int Head[N*11],tot;
26 int n,m,k,s,t,ans=INF;
27 inline void Addedge(int x,int y,int z){G[++tot].to=y,G[tot].nxt=Head[x],Head[x]=tot,G[tot].w=z;}
28 int dis[N*11];
29 priority_queue<pii,vector<pii>,greater<pii> > q;
30 #define y G[j].to
31 inline void dij(){
32     memset(dis,0x3f,sizeof dis);q.push(make_pair(dis[k*n+s]=0,k*n+s));
33     while(!q.empty()){
34         int x=q.top().second,d=q.top().first;q.pop();
35         if(d^dis[x])continue;
36         for(register int j=Head[x];j;j=G[j].nxt)if(MIN(dis[y],d+G[j].w))q.push(make_pair(dis[y],y));
37     }
38 }
39 #undef y
40 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
41     read(n),read(m),read(k),read(s),read(t);++s,++t;
42     for(register int i=1,x,y,z;i<=m;++i){
43         read(x),read(y),read(z);++x,++y;
44         for(register int j=0,l=0;j<=k;++j,l+=n)Addedge(l+x,l+y,z),Addedge(l+y,l+x,z);
45         for(register int j=0,l=0;j<k;++j,l+=n)Addedge(l+n+x,l+y,0),Addedge(l+n+y,l+x,0);
46     }
47     dij();
48     for(register int i=0;i<=k;++i)MIN(ans,dis[i*n+t]);
49     return printf("%d\n",ans),0;
50 }

注：其实这么多边没必要建出来，跑dij的时候直接按原图的边，做免费和不免费两种转移就行了。

标签：return,JLOI2011,int,短路,tot,BZOJ2763,read,include,dis
来源： https://www.cnblogs.com/saigyouji-yuyuko/p/11607052.html