分层最短路（牛客第四场）-- free

题意：

给你边权，起点和终点，有k次机会把某条路变为0，问你最短路是多长。

思路：

分层最短路模板题。题目有点坑（卡掉了SPFA，只能用dijkstra跑的算法）。

 1 #include
 2 #include
 3 #include
 4 #include
 5 using namespace std;
 6  
 7 const int MAXN=1e3,MAXM=5e3,MAXK=1000,INF=~0U>>1;
 8 int N,M,K;
 9 int S,T;
10  
11 struct E{int to,next,value;} e[4*MAXM*(MAXK+1)+MAXK+1];int ecnt,G[MAXN*(MAXK+1)+1];
12 void addEdge(int u,int v,int w){e[++ecnt]={v,G[u],w};G[u]=ecnt;}
13  
14 void readInt(int & x)
15 {
16     x=0;
17     bool flag=false;
18     char c;
19     do c=getchar(); while(c!='-'&&(c<'0'||c>'9'));
20     if(c=='-') flag=true;
21     else x+=c-'0';
22     c=getchar();
23     while(c>='0'&&c<='9')
24     {
25         x*=10;
26         x+=c-'0';
27         c=getchar();
28     }
29 }
30  
31 inline void read()
32 {
33     int i,j;
34     readInt(N);readInt(M);
35     readInt(S);readInt(T);readInt(K);
36     for(i=1;i<=M;i++)
37     {
38         int u,v,w;
39         readInt(u);readInt(v);readInt(w);
40         for(j=0;j<=K;j++)
41         {
42             addEdge(u+N*j,v+N*j,w);
43             addEdge(v+N*j,u+N*j,w);
44             if(j!=K)
45             {
46                 addEdge(u+N*j,v+N*(j+1),0);
47                 addEdge(v+N*j,u+N*(j+1),0);
48             }
49         }
50     }
51 }
52  
53 struct HN{int id,v;bool operator<(const HN & ot)const{return v>ot.v;}};
54  
55 priority_queue heap;
56  
57 bool inS[MAXN*(MAXK+1)+1];
58 int dis[MAXN*(MAXK+1)+1];
59  
60 void dijkstra(int v0)
61 {
62     int i;
63     for(i=1;i<=N*(K+1);i++) dis[i]=INF;
64     memset(inS,false,sizeof(inS));
65     dis[v0]=0;
66     heap.push((HN){v0,0});
67     while(!heap.empty())
68     {
69         int u=heap.top().id;heap.pop();
70         if(inS[u]) continue;
71         inS[u]=true;
72         for(i=G[u];i;i=e[i].next)
73         {
74             int v=e[i].to;
75             if(!inS[v])
76                 if(dis[v]>dis[u]+e[i].value)
77                 {
78                     dis[v]=dis[u]+e[i].value;
79                     heap.push((HN){v,dis[v]});
80                 }
81         }
82     }
83 }
84  
85 int main()
86 {
87     //freopen("Motor.in","r",stdin);
88     //freopen("Motor.out","w",stdout);
89     int i;
90     read();
91     dijkstra(S);
92     int ans=INF;
93     for(i=0;i<=K;i++) ans=min(ans,dis[T+N*i]);
94     printf("%d",ans);
95     return 0;
96 }

 

转载于:https://www.cnblogs.com/--HPY-7m/p/11437347.html