Adding some more judges, here and there.
[and.git] / UVa / 820 - Internet bandwidth / 820.2.cpp
blob5ff5e40c4f047392272bf3bb4c8f481484bde5b2
1 /*
2 Ensayando una implementación un poco diferente.
3 Encuentra el flujo máximo pero no permite reconstruir
4 la red de flujos, i.e, no sé cuanto material fluye de
5 i a j.
7 Accepted.
8 */
9 #include <iostream>
10 #include <iterator>
11 #include <queue>
12 #include <stdio.h>
14 using namespace std;
16 const int MAXN = 101;
18 int cap[MAXN+1][MAXN+1], prev[MAXN+1];
20 int fordFulkerson(int n, int s, int t){
21 int ans = 0;
22 while (true){
23 fill(prev, prev+n, -1);
24 queue<int> q;
25 q.push(s);
26 while (q.size() && prev[t] == -1){
27 int u = q.front();
28 q.pop();
29 for (int v = 0; v<n; ++v)
30 if (v != s && prev[v] == -1 && cap[u][v] > 0)
31 prev[v] = u, q.push(v);
34 if (prev[t] == -1) break;
36 int bottleneck = INT_MAX;
37 for (int v = t, u = prev[v]; u != -1; v = u, u = prev[v]){
38 bottleneck = min(bottleneck, cap[u][v]);
40 for (int v = t, u = prev[v]; u != -1; v = u, u = prev[v]){
41 cap[u][v] -= bottleneck;
42 cap[v][u] += bottleneck;
44 ans += bottleneck;
46 return ans;
49 int main(){
50 int n, runs = 1;
51 while (scanf("%d", &n) == 1 && n){
52 for (int i=0; i<n; ++i){
53 fill(cap[i], cap[i]+n, 0);
55 int s, t, C;
56 scanf("%d %d %d", &s, &t, &C);
57 --s, --t;
58 while (C--){
59 int i, j, k;
60 scanf("%d %d %d", &i, &j, &k);
61 cap[--i][--j] += k, cap[j][i] += k;
64 printf("Network %d\nThe bandwidth is %d.\n\n", runs++, fordFulkerson(n, s, t));
66 return 0;