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問題編輯,現在我只想知道是否可以使用隊列來改進算法。優化dijkstra實現
我發現這個實現混合代價最大流算法,它採用的Dijkstra:http://www.stanford.edu/~liszt90/acm/notebook.html#file2
要去這裏粘貼的情況下,它就會在互聯網虛空丟失:
// Implementation of min cost max flow algorithm using adjacency
// matrix (Edmonds and Karp 1972). This implementation keeps track of
// forward and reverse edges separately (so you can set cap[i][j] !=
// cap[j][i]). For a regular max flow, set all edge costs to 0.
//
// Running time, O(|V|^2) cost per augmentation
// max flow: O(|V|^3) augmentations
// min cost max flow: O(|V|^4 * MAX_EDGE_COST) augmentations
//
// INPUT:
// - graph, constructed using AddEdge()
// - source
// - sink
//
// OUTPUT:
// - (maximum flow value, minimum cost value)
// - To obtain the actual flow, look at positive values only.
#include <cmath>
#include <vector>
#include <iostream>
using namespace std;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long L;
typedef vector<L> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
const L INF = numeric_limits<L>::max()/4;
struct MinCostMaxFlow {
int N;
VVL cap, flow, cost;
VI found;
VL dist, pi, width;
VPII dad;
MinCostMaxFlow(int N) :
N(N), cap(N, VL(N)), flow(N, VL(N)), cost(N, VL(N)),
found(N), dist(N), pi(N), width(N), dad(N) {}
void AddEdge(int from, int to, L cap, L cost) {
this->cap[from][to] = cap;
this->cost[from][to] = cost;
}
void Relax(int s, int k, L cap, L cost, int dir) {
L val = dist[s] + pi[s] - pi[k] + cost;
if (cap && val < dist[k]) {
dist[k] = val;
dad[k] = make_pair(s, dir);
width[k] = min(cap, width[s]);
}
}
L Dijkstra(int s, int t) {
fill(found.begin(), found.end(), false);
fill(dist.begin(), dist.end(), INF);
fill(width.begin(), width.end(), 0);
dist[s] = 0;
width[s] = INF;
while (s != -1) {
int best = -1;
found[s] = true;
for (int k = 0; k < N; k++) {
if (found[k]) continue;
Relax(s, k, cap[s][k] - flow[s][k], cost[s][k], 1);
Relax(s, k, flow[k][s], -cost[k][s], -1);
if (best == -1 || dist[k] < dist[best]) best = k;
}
s = best;
}
for (int k = 0; k < N; k++)
pi[k] = min(pi[k] + dist[k], INF);
return width[t];
}
pair<L, L> GetMaxFlow(int s, int t) {
L totflow = 0, totcost = 0;
while (L amt = Dijkstra(s, t)) {
totflow += amt;
for (int x = t; x != s; x = dad[x].first) {
if (dad[x].second == 1) {
flow[dad[x].first][x] += amt;
totcost += amt * cost[dad[x].first][x];
} else {
flow[x][dad[x].first] -= amt;
totcost -= amt * cost[x][dad[x].first];
}
}
}
return make_pair(totflow, totcost);
}
};
我的問題是否可以通過使用Dijkstra()中的優先級隊列來改進。我試過,但我無法讓它正常工作。 其實我懷疑在Dijkstra它應該在相鄰節點上循環,而不是所有節點...
非常感謝。
1)目前尚不清楚你要完成什麼2)標題與算法不匹配。這是最低成本還是最高流量? – Haile
要重新說明我的問題。 – Papipo