POJ - 1511 Invitation Cards
Posted 西北会法语
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In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.
The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where ‘X‘ denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.
All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.
Input
Output
Sample Input
2 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50
Sample Output
46 210
输入量很大可以将dj进行堆优化,详细见代码。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #include<algorithm> 6 #define INF 0x3f3f3f3f3f3f3f3f 7 #define maxn 1000010 8 9 using namespace std; 10 11 struct EDGE 12 { 13 int u,v,w,next; 14 }edge[maxn]; 15 16 struct node 17 { 18 int u; 19 int d; 20 node(int u,int d):u(u),d(d) {} 21 bool operator <(const node& p)const 22 { 23 return d>p.d; 24 } 25 }; 26 27 long long dis[maxn]; 28 int a[maxn][3],head[maxn]; 29 bool vis[maxn]; 30 int cnt,n,m; 31 32 void Init() 33 { 34 cnt=0; 35 for(int i=1;i<=n;i++) 36 { 37 vis[i]=false; 38 dis[i]=INF; 39 head[i]=-1; 40 } 41 } 42 43 void addedge(int u,int v,int w) 44 { 45 edge[cnt].u=u; 46 edge[cnt].v=v; 47 edge[cnt].w=w; 48 edge[cnt].next=head[u]; 49 head[u]=cnt++; 50 } 51 52 long long dj(int s) 53 { 54 priority_queue<node> p; 55 dis[s]=0; 56 p.push(node(1,0)); 57 while(!p.empty()) 58 { 59 node p1=p.top(); 60 p.pop(); 61 int u=p1.u; 62 if(vis[u]) 63 continue; 64 vis[u]=true; 65 for(int i=head[u];~i;i=edge[i].next) 66 { 67 int v=edge[i].v; 68 int w=edge[i].w; 69 //cout<<"bian"<<i<<" "<<w<<endl; 70 if(!vis[v]&&dis[v]>dis[u]+w) 71 { 72 dis[v]=dis[u]+w; 73 p.push(node(v,dis[v])); 74 //cout<<v<<" "<<dis[v]<<endl; 75 } 76 } 77 } 78 long long ans=0; 79 for(int i=1;i<=n;i++) 80 { 81 ans+=dis[i]; 82 } 83 return ans; 84 } 85 86 int main() 87 { 88 int T; 89 cin>>T; 90 while(T--) 91 { 92 cin>>n>>m; 93 Init(); 94 for(int i=0;i<m;i++) 95 { 96 scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]); 97 addedge(a[i][0],a[i][1],a[i][2]); 98 } 99 long long ans=dj(1); 100 Init(); 101 for(int i=0;i<m;i++) 102 addedge(a[i][1],a[i][0],a[i][2]); 103 ans+=dj(1); 104 cout<<ans<<endl; 105 } 106 107 108 109 return 0; 110 }
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