Network of Byteland consists of n servers, connected by m optical cables. Each cable connects two servers and can transmitdata 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 takecontrol over some cables so that it could see what data istransmitted by the president palace users. Of course they want to control such set of cables, that it is impossible todownload any data from the global network to the president palacewithout 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 somecost. Since the company's main business is not spying, butproviding internet connection to home users, its management wants to make the operation a good investment. So it wants to buy sucha 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.
Input
There are several test cases in the input. The first line of each case contains n and m ( 2 <= n <= 100 , 1 <= m <= 400 ). Next m lines describecables~--- each cable is described with three integer numbers:servers it connects and the cost of the cable. Cost of each cableis positive and does not exceed 107 .Any two servers are connected by at most one cable. Nocable connects a server to itself.The network is guaranteed to be connected, it is possibleto transmit data from any server to any other one.
There is an empty line between each cases.Output
First output k --- the number of cables to buy. After thatoutput the cables to buy themselves. Cables are numbered startingfrom one in order they are given in the input file.There should an empty line between each cases.Example
| 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 |
Source: Andrew Stankevich's Contest #8
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL long long
#define eps 1e-6
#define M 2010
#define N 110
using namespace std;
struct Edge{int x,y;double w;} g[M];
int n,m; LL sum=0;
bool v[N];
namespace ISAP
{
int H[N],d[N],cur[N],pre[N],gap[M],Q[M];
struct Edge{int u,v,n;double c;} E[M];
int nv,head,tail,cntE; double flow,f;
void init(){cntE=0; memset(H,-1,sizeof(H));}
void addedge(int u,int v,double c)
{
E[cntE]=Edge{u,v,H[u],c}; H[u]=cntE++;
E[cntE]=Edge{v,u,H[v],0}; H[v]=cntE++;
}
void revbfs(int s,int t)
{
head=tail=0 ;
memset(d,-1,sizeof(d));
memset(gap,0,sizeof(gap));
Q[tail++]=t;d[t]=0;gap[d[t]]=1;
while (head!=tail)
{
int u=Q[head++];
for (int i=H[u];~i;i=E[i].n)
{
int v=E[i].v; if (~d[v]) continue;
d[v]=d[u]+1; gap[d[v]]++; Q[tail++]=v;
}
}
}
double isap(int s,int t)
{
memcpy(cur,H,sizeof(cur)); nv=t;
flow=0; revbfs(s,t); int u=pre[s]=s,i;
while (d[s]<nv)
{
if (u==t)
{
f=INF;
for (i=s;i!=t;i=E[cur[i]].v)
if (f-E[cur[i]].c>eps) f=E[cur[i]].c,u=i;
flow += f;
for (i=s;i!=t;i=E[cur[i]].v)
E[cur[i]].c-=f,E[cur[i]^1].c+=f;
}
for (i=cur[u];~i;i=E[i].n)
if (fabs(E[i].c)>eps&&d[u]==d[E[i].v]+1) break ;
if (~i) cur[u]=i,pre[E[i].v]=u,u=E[i].v;
else
{
if (0==--gap[d[u]]) break ;
int minv=nv,v;
for (int i=H[u];~i;i=E[i].n)
{
v=E[i].v;
if (fabs(E[i].c)>eps&&minv>d[v]) minv=d[v],cur[u]=i;
}
d[u]=minv+1; gap[d[u]]++; u=pre[u];
}
}
return flow ;
}
void dfs(int x)
{
v[x]=1;
for(int i=H[x];~i;i=E[i].n)
{
int to=E[i].v;
if (v[to]||E[i].c<eps) continue;
dfs(to);
}
}
}
bool check(double x)
{
using namespace ISAP;
init();
double ans=0;
for(int i=1;i<=m;i++)
if (g[i].w-x>eps)
{
addedge(g[i].x,g[i].y,g[i].w-x);
addedge(g[i].y,g[i].x,g[i].w-x);
} else ans+=g[i].w-x;
ans+=isap(1,n);
return ans>eps;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
sum=0;
memset(g,0,sizeof(g));
for(int i=1;i<=m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
sum+=w;g[i]=Edge{u,v,w};
}
double l=0,r=sum,mid;
while(r-l>eps)
{
mid=(r+l)/2.0;
if (check(mid)) l=mid;
else r=mid;
}
memset(v,0,sizeof(v));
ISAP::dfs(1); vector<int> ans;
for(int i=1;i<=m;i++)
if (v[g[i].x]+v[g[i].y]==1||g[i].w-mid<eps) ans.push_back(i);
printf("%d\n",ans.size());
printf("%d",ans[ans.size()-1]);
for(int i=0;i<(int)ans.size()-1;i++)
printf(" %d",ans[i]);
puts("\n");
}
return 0;
}