SPFA forfeit recursive optimal ratio cycloalkyl ring + || [Usaco2007 Dec] Cow's travel || BZOJ 1690 || Luogu P2868

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Questions surface: P2868 [USACO07DEC] sightseeing Sightseeing Cows Cows

Solution: the optimal ratio of ring

The actual subject is a requirement ans, such that for any one ring satisfying FIG sig (i = 1, n) v [i] / sig (i = 1, n) e [i] <= ans

Therefore, the formula is converted into: sig (i = 1, n) v [i] - [(sig (i = 1, n) v [i]) * ans] <= 0

sig(i=1,n)(v[i]-ans*e[i])<=0

Final into: sig (i = 1, n) (ans * e [i] -v [i])> = 0, i.e. ans * e [i] -v [i] to a new right side in FIG reconstructed, FIG respect to any one of the rings can satisfy this condition shall ans

Therefore binary answer, for each reconstructed FIG mid, SPFA a recursive loop negative judgment, if the sig (i = 1, n) (mid * e [i] -v [i]) <0 then the mid <ans, or DESCRIPTION mid> = ans

Code:

 

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=1050,maxm=5050,inf=(1<<30)-5;
 5 const double jd=1e-3;
 6 int N,M,num_edge=0,edge_head[maxn],u,v;
 7 double V[maxn],e,l,r,mid,Dis[maxn];
 8 bool vis[maxn],flag;
 9 struct Edge{ int to,nx;double e,dis; }edge[maxm];
10 inline void Add_edge(int from,int to,double e){
11     edge[++num_edge].nx=edge_head[from];
12     edge[num_edge].to=to;
13     edge[num_edge].e=e;
14     edge_head[from]=num_edge;
15     return;
16 }
17 inline void Rebuild(){
18     for(int i=1;i<=N;i++)
19         for(int j=edge_head[i];j;j=edge[j].nx)
20             edge[j].dis=edge[j].e*mid-V[edge[j].to];
21     return;
22 }
23 inline void SPFA(int x){
24     if(flag) return;
25     vis[x]=1;
26     for(int i=edge_head[x];i;i=edge[i].nx){
27         int y=edge[i].to;
28         if(Dis[y]>Dis[x]+edge[i].dis){
29             if(vis[y]){
30                 flag=1;
31                 return;
32             }
33             Dis[y]=Dis[x]+edge[i].dis;
34             SPFA(y);
35         }
36     }
37     vis[x]=0;
38 }
39 inline bool Check(){
40     for(int i=1;i<=N;i++) vis[i]=0,Dis[i]=0;
41     flag=0;
42     for(int i=1;i<=N;i++){
43         SPFA(i);
44         if(flag) break;
45     }
46     return flag;
47 }
48 int main(){
49     scanf("%d%d",&N,&M);
50     for(int i=1;i<=N;i++) scanf("%lf",&V[i]);
51     for(int i=1;i<=M;i++){
52         scanf("%d%d%lf",&u,&v,&e);
53         Add_edge(u,v,e);
54     }
55     l=0; r=10000;
56     while(l+jd<r){
57         mid=(l+r)/2;
58         Rebuild();
59         if(Check()) l=mid;
60         else r=mid;
61     }
62     printf("%.2lf\n",l);
63     return 0;
64 }

View Code

 

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By:AlenaNuna