Cod sursa(job #1362808)

Utilizator mirceadinoMircea Popoveniuc mirceadino Data 26 februarie 2015 15:41:00
Problema Cuplaj maxim de cost minim Scor 10
Compilator cpp Status done
Runda Arhiva educationala Marime 2.47 kb
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<deque>

using namespace std;

#ifdef HOME
const string inputFile = "input.txt";
const string outputFile = "output.txt";
#else
const string problemName = "cmcm";
const string inputFile = problemName + ".in";
const string outputFile = problemName + ".out";
#endif

const int INF = (1 << 30);
const int NMAX = 300 * 2 + 5;

int N, M, E;
int C[NMAX][NMAX];
int F[NMAX][NMAX];
int P[NMAX][NMAX];
int Q[NMAX][NMAX];
int D[NMAX];
int dad[NMAX];
vector<int> V[NMAX];
bitset<NMAX> viz;
int source, sink;
int maxflow, cost;
deque<int> DQ;

int bf(int x) {
    int i;

    viz = 0;
    DQ.resize(0);

    for(i = source; i <= sink; i++)
        D[i] = INF;

    D[x] = 0;
    viz[x] = 1;
    DQ.push_back(x);

    while(!DQ.empty()) {
        x = DQ.front();
        DQ.pop_front();

        viz[x] = 0;

        if(x == sink)
            return 1;

        for(auto y : V[x])
            if(F[x][y] < C[x][y] && D[x] + P[x][y] < D[y]) {
                D[y] = D[x] + P[x][y];
                dad[y] = x;

                if(!viz[y]) {
                    viz[y] = 1;
                    DQ.push_back(y);
                }
            }
    }

    return 0;
}

int main() {
    int i, j, x, y, z, v;

    freopen(inputFile.c_str(), "r", stdin);
    freopen(outputFile.c_str(), "w", stdout);

    scanf("%d%d%d", &N, &M, &E);

    for(i = 1; i <= E; i++) {
        scanf("%d%d%d", &x, &y, &z);
        C[x][N + y] = INF;
        P[x][N + y] = z;
        P[N + y][x] = -z;
        Q[x][y] = i;
        V[x].push_back(N + y);
        V[N + y].push_back(x);
    }

    source = 0;
    sink = N + M + 1;

    for(i = 1; i <= N; i++) {
        V[source].push_back(i);
        V[i].push_back(source);
        C[source][i] = 1;
    }

    for(i = 1; i <= M; i++) {
        V[N + i].push_back(sink);
        V[sink].push_back(N + i);
        C[N + i][sink] = 1;
    }

    while(bf(source)) {
        v = INF;

        for(x = sink; x != source; x = dad[x])
            v = min(v, C[dad[x]][x] - F[dad[x]][x]);

        for(x = sink; x != source; x = dad[x]) {
            F[dad[x]][x] += v;
            F[x][dad[x]] -= v;
        }

        maxflow += v;
    }

    for(i = 1; i <= N; i++)
        for(j = 1; j <= M; j++)
            if(F[i][N + j])
                cost += P[i][N + j];

    printf("%d %d\n", maxflow, cost);

    for(i = 1; i <= N; i++)
        for(j = 1; j <= M; j++)
            if(F[i][N + j])
                printf("%d ", Q[i][j]);

    return 0;
}