bzoj 4501 旅行

01分数规划+最大权闭合子图 
倒拓扑序处理每个节点 

$$f[x]=\frac{\sum{f[v]}}{n}+1$$

二分答案$val$

只需要判断是否存在$\sum{f[v]}+1-val>0$即可 
点权下放给边，限制${x,y}$即为若边$x$存在，则边$y$存在 
建图，跑网络流即可

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<cmath>
  6 #include<vector>
  7 #define inf 0x7fffffff
  8 #define eps 1e-7
  9 using namespace std;
 10 
 11 
 12 namespace network{
 13     const int N = 1005;
 14     const int S = 1001;
 15     const int T = 1002;
 16     int e=2,head[N];
 17     struct edge{
 18         int u,v,next;
 19         double f;
 20     }ed[310000];
 21     void add(int u,int v,double f){
 22         ed[e].u=u;ed[e].v=v;ed[e].f=f;
 23         ed[e].next=head[u];head[u]=e++;
 24         ed[e].u=v;ed[e].v=u;ed[e].f=0;
 25         ed[e].next=head[v];head[v]=e++;
 26     }
 27     int dep[N];
 28     int q[N],h,t;
 29     bool bfs(){
 30         memset(dep,-1,sizeof dep);
 31         dep[S]=1; h=t=1; q[1]=S;
 32         while(h<=t){
 33             int x=q[h++];
 34             for(int i=head[x];i;i=ed[i].next){
 35                 if(ed[i].f&&dep[ed[i].v]==-1){
 36                     dep[ed[i].v]=dep[x]+1;
 37                     q[++t]=ed[i].v;
 38                 }
 39             }
 40         }
 41         return dep[T]!=-1;
 42     }
 43     double dfs(int x,double f){
 44         if(x==T||f==0)return f;
 45         double ans=0;
 46         for(int i=head[x];i;i=ed[i].next){
 47             if(ed[i].f&&dep[ed[i].v]==dep[x]+1){
 48                 double nxt=dfs(ed[i].v,min(f,ed[i].f));
 49                 ed[i].f-=nxt; ed[i^1].f+=nxt;
 50                 f-=nxt; ans+=nxt;
 51                 if(f==0)break;
 52             }
 53         }
 54         if(ans==0)dep[x]=-1;
 55         return ans;
 56     }
 57     double dinic(){
 58         double ans=0;
 59         while(bfs())ans+=dfs(S,inf);
 60         return ans;
 61     }
 62     void init(){memset(head,0,sizeof head);e=2;}
 63 }
 64 
 65 
 66 namespace graph{
 67     const int N =505;
 68     int e=1,head[N],out[N];
 69     vector<int> lim[N];
 70     bool vis[N];
 71     double f[N];
 72     struct edge{
 73         int u,v,next;
 74     }ed[2*N];
 75     void add(int u,int v){
 76         ed[e].u=u;ed[e].v=v;
 77         ed[e].next=head[u];
 78         head[u]=e++;out[u]++;
 79     }
 80     bool check(int x,double y){
 81         network::init();
 82         double sum=0;
 83         for(int i=head[x];i;i=ed[i].next){
 84             int v=ed[i].v;
 85             double val=f[v]+1-y;
 86             sum+=max(val,(double)0);
 87             if(val>0)network::add(network::S,i,val);
 88             else network::add(i,network::T,-val);
 89             for(int j=0;j<lim[i].size();j++)
 90                 network::add(i,lim[i][j],(double)inf);
 91         }
 92         return sum-network::dinic()>0;
 93     }
 94     double solve(int x){
 95         if(!out[x])return 0.0;
 96         double l=0,r=0,mid,ans;
 97         for(int i=head[x];i;i=ed[i].next)
 98             r=max(r,f[ed[i].v]+1);
 99         while(r-l>=eps){
100             mid=(l+r)/2.0;
101             if(check(x,mid))l=ans=mid;
102             else r=mid;
103         }
104         return ans;
105     }
106     void work(int x){
107         for(int i=head[x];i;i=ed[i].next)
108             if(!vis[ed[i].v]){
109                 work(ed[i].v);
110                 vis[ed[i].v]=1;
111             }
112         f[x]=solve(x);
113     }
114 }
115 
116 int main(){
117     int n,m,k;
118     scanf("%d%d%d",&n,&m,&k);
119     for(int i=1,u,v;i<=m;i++){
120         scanf("%d%d",&u,&v);
121         graph::add(u,v);
122     }
123     for(int i=1,x,y;i<=k;i++){
124         scanf("%d%d",&x,&y);
125         graph::lim[x].push_back(y);
126     }
127     graph::work(1);
128     printf("%lf\n",graph::f[1]);
129 }
130 /*
131 3 3 1
132 1 2
133 1 3
134 2 3
135 1 2
136 */
View Code

 

posted @ 2017-09-26 19:53  Ren_Ivan  阅读(143)  评论(0)  编辑  收藏  举报