Cod sursa(job #2691223)

Utilizator RNedelcuNedelcu Radu RNedelcu Data 27 decembrie 2020 16:35:16
Problema Critice Scor 20
Compilator cpp-64 Status done
Runda Arhiva de probleme Marime 2.19 kb
#include <iostream>
#include <bits/stdc++.h>
#define MAX 1001
using namespace std;
ifstream in("critice.in");
ofstream out("critice.out");
int N,M,S,F; // Nr noduri, start, finish
vector<int> Graph[MAX];
int capacitate[MAX][MAX];
int flow[MAX][MAX];
int muchii[MAX][MAX];

int bfs(vector<int> &tata)
{
    fill(tata.begin(),tata.end(),-1);
    pair<int,int> que[MAX]; //Nod, flow
    int L=1;
    que[L]= {S,2e9};
    tata[S]=S;
    for(int i=1; i<=L; i++)
    {
        int cur = que[i].first;
        int curFlow = que[i].second;
        for(auto next : Graph[cur])
        {
            if(tata[next] == -1 && (capacitate[cur][next]-flow[cur][next]>0)) // Daca nu e deja vizitat si mai putem creste fluxul
            {
                tata[next] = cur;
                int new_flow = min(curFlow,capacitate[cur][next]-flow[cur][next]);
                if(next==F)
                    return new_flow;
                que[++L]= {next,new_flow};
            }
        }
    }
    return 0;
}

int main()
{
    int x,y,z;
    in>>N>>M;
    S=1;
    F=N;
    for(int i=1; i<=M; i++)
    {
        in>>x>>y>>z; //Nod, Nod, capacitatea maxima
        Graph[x].push_back(y);
        Graph[y].push_back(x);
        capacitate[x][y] =capacitate[y][x]= z;
        muchii[x][y]=muchii[y][x]=i;
    }

    vector<int> tata(N+1);
    int new_flow,previous, maxflow=0;
    while(new_flow = bfs(tata)) //Cat timp gasim un lant nesaturat
    {
        maxflow+=new_flow; //Crestem fluxul
        int current = F;
        while(current!=S) //Updatam capacitatile
        {
            previous = tata[current];
            flow[previous][current]+=new_flow; // Pe muchiile directe adun fluxul adaugat
            flow[current][previous]-=new_flow; // Pe muchiile inverse scad
            current = previous;
        }
    }
    vector<int> rez;
    rez.reserve(M);
    for(int i=1; i<=N; i++)
        for(auto j : Graph[i])
            if(tata[i]!=-1 && capacitate[i][j] ) //Daca avem muchie dinspre un nod din taietura spre restul, o afisam
                rez.push_back(muchii[i][j]);
    out<<rez.size()<<'\n';
    for(auto temp: rez)
        out<<temp<<'\n';

    return 0;
}