Sightseeing Cows-POJ3621——01分数规划，SPFA判负环

http://poj.org/problem?id=3621

　　01规划，二分答案，重组边权，SPFA判负环。

code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 using namespace std;
 5 const int N=1000+10;
 6 const int M=5000+10;
 7 const double INF=1e10;
 8 int head[N],ver[M],from[M],nex[M],T[M],F[M],tot=0;
 9 double edge[M];
10 inline void add(int x,int y,int t,int f)
11 {
12     ver[++tot]=y;
13     from[tot]=x;
14     T[tot]=t;
15     F[tot]=f;
16     nex[tot]=head[x];
17     head[x]=tot;
18 }
19 
20 int n,m;
21 int fun[N];
22 
23 bool vis[N];
24 double dis[N];
25 int cnt[N];
26 
27 bool judge(double mi)
28 {
29     memset(vis,0,sizeof(vis));
30     cnt[1]=0;
31     for(int i=1;i<=tot;++i)
32         edge[i]=mi*T[i]-F[i];
33     for(int i=1;i<=n;++i)
34         dis[i]=INF;
35     dis[1]=0;
36     vis[1]=1;
37     queue<int> q;
38     q.push(1);
39 
40     while(!q.empty())
41     {
42         int x=q.front();
43         q.pop();
44         vis[x]=0;
45         for(int i=head[x];i;i=nex[i])
46         {
47             int y=ver[i];
48             if(dis[y]>dis[x]+edge[i])
49             {
50                 dis[y]=dis[x]+edge[i];
51                 cnt[y]=cnt[x]+1;
52                 if(cnt[y]>=n)
53                     return true;
54                 if(vis[y])continue;
55                 vis[y]=1;
56                 q.push(y);
57             }
58         }
59     }
60     return false;
61 }
62 
63 double eps=1e-8;
64 void solve()
65 {
66     double l=0,r=1100;
67     while(r-l>eps)
68     {
69         double m=(l+r)/2;
70         if(judge(m))
71             l=m;
72         else r=m;
73     }
74     if(l<eps)
75         puts("0");
76     else printf("%.2f\n",l);
77 }
78 
79 int main()
80 {
81     scanf("%d%d",&n,&m);
82     for(int i=1;i<=n;++i)
83         scanf("%d",fun+i);
84     for(int i=0;i<m;++i)
85     {
86         int x,y,t;
87         scanf("%d%d%d",&x,&y,&t);
88         add(x,y,t,fun[x]);
89     }
90     solve();
91     return 0;
92 }

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