Pagini recente » Cod sursa (job #274280) | Cod sursa (job #947492) | Cod sursa (job #633777) | Cod sursa (job #765702) | Cod sursa (job #1292905)
#include <fstream>
#include <vector>
#include <string.h>
using namespace std;
ifstream ka("cmcm.in");
ofstream ki("cmcm.out");
#define MAX_N 710
#define inf 2000000000
int n, m, e, dest, sol, nr, improve;
int Q[MAX_N * MAX_N], tata[MAX_N], dist[MAX_N], use[MAX_N];
int C[MAX_N][MAX_N], F[MAX_N][MAX_N], edge[MAX_N][MAX_N];
vector <int> v[MAX_N], cost[MAX_N];
void build_graph() {
//sursa este nodul 1, destinatia n + m + 2
dest = n + m + 2;
for (int i = 2; i <= n + 1; i++)
{
v[1].push_back(i);
cost[1].push_back(0);
v[i].push_back(1);
cost[i].push_back(0);
C[1][i] = 1;
}
for (int i = n + 2; i <= n + m + 1; i++) {
v[i].push_back(dest);
cost[i].push_back(0);
v[dest].push_back(i);
cost[dest].push_back(0);
C[i][dest] = 1;
}
}
int bellman_ford() {
for (int i = 1; i <= dest; i++)
{
dist[i] = inf;
tata[i] = -1;
use[i] = 0;
}
dist[1] = 0;
use[1] = 1;
int st = 0, dr = 1;
Q[1] = 1;
while (st < dr)
{
st++;
int len = v[Q[st]].size();
for (int i = 0; i < len; i++)
{
if (C[Q[st]][v[Q[st]][i]] > F[Q[st]][v[Q[st]][i]] && dist[v[Q[st]][i]] > dist[Q[st]] + cost[Q[st]][i]) {
dist[v[Q[st]][i]] = dist[Q[st]] + cost[Q[st]][i];
tata[v[Q[st]][i]] = Q[st];
if (!use[v[Q[st]][i]])
{
Q[++dr] = v[Q[st]][i];
use[v[Q[st]][i]] = 1;
}
}
}
use[Q[st]] = 0;
}
if (dist[dest] < inf)
{
int flux = inf;
for (int i = dest; i != 1; i = tata[i])
flux = min(flux, C[tata[i]][i] - F[tata[i]][i]);
for (int i = dest; i != 1; i = tata[i])
{
F[tata[i]][i] += flux;
F[i][tata[i]] -= flux;
}
return flux * dist[dest];
}
return 0;
}
int main()
{
ka >> n >> m >> e;
for (int i = 1; i <= e; i++)
{
int p, q, val;
ka >> p >> q >> val;
p++;
q += n + 1;
v[p].push_back(q);
cost[p].push_back(val);
v[q].push_back(p);
cost[q].push_back(-val);
edge[p][q] = i;
C[p][q] = 1;
}
build_graph();
improve = 1;
while (improve)
{
improve = bellman_ford();
sol += improve;
}
for (int i = 2; i <= n + 1; i++)
for (int j = n + 2; j < dest; j++)
if (C[i][j] && F[i][j])
{
nr++;
break;
}
ki << nr << " " << sol << '\n';
for (int i = 2; i <= n + 1; i++)
for (int j = n + 2; j < dest; j++)
if (C[i][j] && F[i][j])
{
ki << edge[i][j] << " ";
break;
}
}