Cod sursa(job #768222)

Utilizator iuli1505Parasca Iuliana iuli1505 Data 16 iulie 2012 13:21:38
Problema Cuplaj maxim de cost minim Scor 100
Compilator cpp Status done
Runda Arhiva educationala Marime 2.54 kb
#include<cstdio>
#include<vector>
#include<deque>
#define nmax 1000
#define oo 1<<30
using namespace std;
void read(),solve();
int bf();
int n,m,e,sursa,dest;
int cap[nmax][nmax],M[nmax][nmax],flux[nmax][nmax],cost[nmax][nmax];
int d[nmax],viz[nmax],dad[nmax];
vector<int>v[nmax];
deque<int>Q;
int main()
{
    read();
    solve();
    return 0;
}
void read()
{
    int x,y,z,i;
    freopen("cmcm.in","r",stdin);
    freopen("cmcm.out","w",stdout);
    scanf("%d %d %d ", &n, &m, &e);
    for(i=1;i<=e;i++)
    {
        scanf("%d %d %d ", &x, &y, &z);
        y+=n;
        v[x].push_back(y);
        v[y].push_back(x);
        cap[x][y]=1;
        M[x][y]=i;
        cost[x][y]=z;
        cost[y][x]=-z;
    }
    sursa=0;
    dest=n+m+1;
    for(i=1;i<=n;i++)
    {
        v[sursa].push_back(i);
        v[i].push_back(sursa);
        cap[sursa][i]=1;
    }
    for(i=n+1;i<=n+m;i++)
    {
        v[i].push_back(dest);
        v[dest].push_back(i);
        cap[i][dest]=1;
    }
}
void solve()
{
    int i,j;
    int sol=0,nr=0,ok;
    ok=bf();
    for(;ok;)
    {
        sol+=ok;
        ok=bf();
    }
    for(i=1;i<=n;i++)
        for(j=n+1;j<=n+m;j++)
            if(cap[i][j]&&flux[i][j])
            {
                nr++;
                break;
            }
    printf("%d %d\n", nr, sol);
    for(i=1;i<=n;i++)
        for(j=n+1;j<=n+m;j++)
            if(cap[i][j]&&flux[i][j])
                printf("%d ", M[i][j]);
}
int bf()
{
    int i,x,mflux;
    for(i=sursa;i<=dest;i++)
    {
        d[i]=oo;
        dad[i]=-1;
        viz[i]=0;
    }
    d[sursa]=0;
    viz[sursa]=1;
    Q.push_back(sursa);
    for(;Q.size();)
    {
        x=Q.front();
        for(vector<int>::iterator it=v[x].begin();it!=v[x].end();it++)
        {
            if((d[x]+cost[x][*it]<d[*it])&&(cap[x][*it]>flux[x][*it]))
            {
                d[*it]=d[x]+cost[x][*it];
                dad[*it]=x;
                if(!viz[*it])
                {
                    viz[*it]=1;
                    Q.push_back(*it);
                }
            }
        }
        viz[x]=0;
        Q.pop_front();
    }
    if(d[dest]<oo)
    {
        mflux=oo;
        for(i=dest;i!=sursa;i=dad[i])
            if((cap[dad[i]][i]-flux[dad[i]][i])<mflux)
                mflux=cap[dad[i]][i]-flux[dad[i]][i];
        for(i=dest;i!=sursa;i=dad[i])
        {
            flux[dad[i]][i]+=mflux;
            flux[i][dad[i]]-=mflux;
        }
        return mflux*d[dest];
    }
    return 0;
}