ZOJ 2676 Network Wars[01分数规划]

 

ZOJ Problem Set - 2676

Network Wars

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Time Limit: 5 Seconds      Memory Limit: 32768 KB      Special Judge

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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.

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 describe cables~--- each cable is described with three integer numbers: servers it connects and the cost of the cable. Cost of each cable is positive and does not exceed 10⁷.

 

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.

Output

First output k --- the number of cables to buy. After that output the cables to buy themselves. Cables are numbered starting from 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

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Source: Andrew Stankevich's Contest #8
Submit    Status  

include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef double real;
const real eps=1e-6;
const int Z=1e3+5,N=1e4+5,M=1e5+5;
struct data{int u,v,w;}a[Z];
struct edge{int v,next;real cap;}e[M];int tot=1,head[N];
int n,m,S,T,cas,num,g[Z],dis[N],q[M];bool vis[Z];
real L,R,ans;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int dcmp(real x){
    if(fabs(x)<eps) return 0;
    return x>0?1:-1;
}
inline void add(int x,int y,real z){
    e[++tot].v=y;e[tot].cap=z;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].cap=z;e[tot].next=head[y];head[y]=tot;
}
inline bool bfs(){
    for(int i=S;i<=T;i++) dis[i]=-1;
    int h=0,t=1;q[t]=S;dis[S]=0;
    while(h!=t){
        int x=q[++h];
        for(int i=head[x];i;i=e[i].next){
            if(dcmp(e[i].cap)>0&&dis[e[i].v]==-1){
                dis[e[i].v]=dis[x]+1;
                if(e[i].v==T) return 1;
                q[++t]=e[i].v;
            }
        }
    }
    return 0;
}
real dfs(int x,real f){
    if(x==T) return f;
    real used=0,t;
    for(int i=head[x];i;i=e[i].next){
        if(dcmp(e[i].cap)>0&&dis[e[i].v]==dis[x]+1){
            t=dfs(e[i].v,min(e[i].cap,f));
            e[i].cap-=t;e[i^1].cap+=t;
            used+=t;f-=t;
            if(!f) return used;
        }
    }
    if(!used) dis[x]=-1;
    return used;
}
inline real dinic(){
    real res=0;
    while(bfs()) res+=dfs(S,2e9);
    return res;
}
inline void init(){
    m=read();L=0;R=0;num=0;S=1;T=n;
    for(int i=1;i<=m;i++) a[i].u=read(),a[i].v=read(),a[i].w=read(),R+=a[i].w;
}
inline real rebuild(real s){
    tot=1;memset(head,0,n+1<<2);
    real tans=0; 
    for(int i=1;i<=m;i++){
        if(a[i].w>s){
            add(a[i].u,a[i].v,a[i].w-s);
        }
        else tans+=a[i].w-s;
    }
    return tans+dinic();
}
inline real binary_search(){
    real mid,now;
    while(dcmp(R-L)>0){
        mid=(L+R)/2;
        now=rebuild(mid);
        if(dcmp(now)>0) L=mid;
        else R=mid;
    }
    return mid;
}
void DFS(int x){
    vis[x]=1;
    for(int i=head[x];i;i=e[i].next){
        if(!vis[e[i].v]&&dcmp(e[i].cap)>0){
            DFS(e[i].v);
        }
    }
}
inline void work(){
    ans=binary_search();
    memset(vis,0,n+1<<2);
    //rebuild(ans);
    DFS(S);
    for(int i=1;i<=m;i++){
        if(vis[a[i].u]+vis[a[i].v]==1||a[i].w<ans){
            g[++num]=i;
        }
    }
    if(cas++) putchar('\n');
    printf("%d\n",num);
    for(int i=1;i<num;i++) printf("%d ",g[i]);
    if(num) printf("%d",g[num]);
    putchar('\n');
}
int main(){
    while(~scanf("%d",&n)) init(),work();
    return 0;
}

 

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6606463.html
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