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Network of Byteland consists of n servers, connected by m optical cables. Each cable connects two servers and can transmit data in both directions. Two servers of the network are especially important --- they are connected to global world network and president palace network respectively.
The server connected to the president palace network has number 1, and the server connected to the global world network has number n.
Recently the company Max Traffic has decided to take control over some cables so that it could see what data is transmitted by the president palace users. Of course they want to control such set of cables, that it is impossible to download any data from the global network to the president palace without transmitting it over at least one of the cables from the set.
To put its plans into practice the company needs to buy corresponding cables from their current owners. Each cable has some cost. Since the company‘s main business is not spying, but providing internet connection to home users, its management wants to make the operation a good investment. So it wants to buy such a set of cables, that cables mean cost} is minimal possible.
That is, if the company buys k cables of the total cost c, it wants to minimize the value of c/k.
Any two servers are connected by at most one cable. No cable connects a server to itself. The network is guaranteed to be connected, it is possible to transmit data from any server to any other one.
There is an empty line between each cases.| Input | Output |
6 8 1 2 3 1 3 3 2 4 2 2 5 2 3 4 2 3 5 2 5 6 3 4 6 3 |
4 3 4 5 6 |
4 5 1 2 2 1 3 2 2 3 1 2 4 2 3 4 2 |
3 1 2 3
|
#include"stdio.h"
#include"string.h"
#include"math.h"
#include"iostream"
#include"queue"
#include"stack"
#include"map"
#include"string"
#define M 409
#define inf 0x3f3f3f3f
#define eps 1e-6
using namespace std;
struct node
{
int u,v,next;
double w;
}edge[M*10];
int t,head[M],work[M],a[M],b[M],c[M],dis[M],belong[M],use[M];
double min(double a,double b)
{
return a<b?a:b;
}
void init()
{
t=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,double w)
{
edge[t].u=u;
edge[t].v=v;
edge[t].w=w;
edge[t].next=head[u];
head[u]=t++;
}
int bfs(int S,int T)
{
queue<int>q;
memset(dis,-1,sizeof(dis));
dis[S]=0;
q.push(S);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];~i;i=edge[i].next)
{
int v=edge[i].v;
if(edge[i].w>eps&&dis[v]==-1)
{
dis[v]=dis[u]+1;
q.push(v);
if(v==T)
return 1;
}
}
}
return 0;
}
double dfs(int cur,double a,int T)
{
if(cur==T)return a;
for(int &i=work[cur];~i;i=edge[i].next)
{
int v=edge[i].v;
if(edge[i].w>eps&&dis[v]==dis[cur]+1)
{
double tt=dfs(v,min(a,edge[i].w),T);
if(tt)
{
edge[i].w-=tt;
edge[i^1].w+=tt;
return tt;
}
}
}
return 0;
}
double Dinic(int S,int T)
{
double ans=0;
while(bfs(S,T))
{
memcpy(work,head,sizeof(head));
while(double tt=dfs(S,inf,T))
ans+=tt;
}
return ans;
}
double fun(int n,int m,double r)
{
init();
double sum=0;
for(int i=1;i<=m;i++)
{
if(c[i]>r)
{
add(a[i],b[i],c[i]-r);
add(b[i],a[i],c[i]-r);
}
else
sum+=c[i]-r;
}
return sum+Dinic(1,n);
}
void DFS(int u)
{
use[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(edge[i].w>eps&&!use[v])
DFS(v);
}
}
int main()
{
int n,m,kk=0;
while(scanf("%d%d",&n,&m)!=-1)
{
double l=0,r=0,mid;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&a[i],&b[i],&c[i]);
r+=c[i];
}
while(r-l>eps)
{
mid=(l+r)/2;
double msg=fun(n,m,mid);
if(msg>eps)
{
l=mid;
}
else
r=mid;
}
fun(n,m,mid);//重新跑一遍网络流，因为最后一次的网络流不一定是最优值mid的网络流
memset(use,0,sizeof(use));
DFS(1);
int num=0;
for(int i=1;i<=m;i++)
{
if(use[a[i]]!=use[b[i]]||c[i]<mid)
belong[num++]=i;
}
printf("%d\n",num);
printf("%d",belong[0]);
for(int i=1;i<num;i++)
printf(" %d",belong[i]);
printf("\n");
if(kk)
printf("\n");
kk++;
}
return 0;
}
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原文地址：http://www.cnblogs.com/mypsq/p/4476425.html
