UVA 1349 Optimal Bus Route Design——最小权完美匹配

UVA 1349 Optimal Bus Route Design——最小权完美匹配

对于边u->v，将u拆分成u1 u2, v拆分成v1 v2，然后u1连接v2，s连接u1，v2连接t，就把整个图转化成了一个二分图，然后用二分图的最小权完美匹配就可以解决问题

#include #include #include #include #include #include using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 500;struct Edge { int from, to, cap, flow, cost; Edge (int u, int v, int c, int f, int w) : from(u), to(v), cap(c), flow(f), cost(w) {}};struct MCMF { int n, m; vector edges; vector G[maxn]; int inq[maxn], d[maxn], p[maxn], a[maxn]; void init(int x) { n = x; edges.clear(); for (int i = 0; i <= n; i++) G[i].clear(); } void addedge(int from, int to, int cap, int cost) { edges.push_back(Edge(from, to, cap, 0, cost)); edges.push_back(Edge(to, from, 0, 0, -cost)); m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } bool BellmanFord(int s, int t, int &flow, long long &cost) { memset(inq, 0, sizeof(inq)); for (int i = 0; i <= n; i++) d[i] = INF; inq[s] = 1, d[s] = 0, p[s] = 0, a[s] = INF; queue q; q.push(s); while (!q.empty()) { int u = q.front(); q.pop(); inq[u] = 0; for (int i = 0; i < G[u].size(); i++) { Edge &e = edges[G[u][i]]; if (e.cap > e.flow && d[e.to] > d[u] + e.cost) { d[e.to] = d[u] + e.cost; p[e.to] = G[u][i]; a[e.to] = min(a[u], e.cap - e.flow); if (!inq[e.to]) { q.push(e.to); inq[e.to] = 1; } } } } if (d[t] == INF) return false; flow += a[t]; cost += (long long)d[t] * (long long)a[t]; for (int u = t; u != s; u = edges[p[u]].from) { edges[p[u]].flow += a[t]; edges[p[u]^1].flow -= a[t]; } return true; } int mincostmaxflow(int s, int t, long long &cost) { int flow = 0; cost = 0; while (BellmanFord(s, t, flow, cost)); return flow; }};int main() { MCMF mcmf; int n; while (~scanf("%d", &n) && n) { mcmf.init(n*2+5); for (int i = 1; i <= n; i++) { int v; while (scanf("%d", &v) && v) { int cost; scanf("%d", &cost); mcmf.addedge(i, v + n, 1, cost); } } int s = 0, t = (n<<1) + 1; for (int i = 1; i <= n; i++) { mcmf.addedge(s, i, 1, 0); mcmf.addedge(i + n, t, 1, 0); } long long cost = 0; if (mcmf.mincostmaxflow(s, t, cost) != n) printf("N\n"); else cout << cost << endl; } return 0;}

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