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UVa1658 Admiral (拆点法,最小费用流)
链接:http://vjudge.net/problem/UVA-1658
分析:把2到v-1的每个节点i拆成i和i‘两个结点,中间连一条容量为1,费用为0的边,然后求1到v的流量为2的最小费用流即可。
1 #include <cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<vector> 5 #include<algorithm> 6 using namespace std; 7 8 const int maxn = 2000 + 5; 9 const int INF = 1000000000;10 11 struct Edge {12 int from, to, cap, flow, cost;13 Edge(int u, int v, int c, int f, int w):from(u), to(v), cap(c), flow(f), cost(w) {}14 };15 16 struct MCMF {17 int n, m;18 vector<Edge> edges;19 vector<int> G[maxn];20 int inq[maxn];21 int d[maxn];22 int p[maxn];23 int a[maxn];24 25 void init(int n) {26 this -> n = n;27 for (int i = 0; i < n; i++) G[i].clear();28 edges.clear();29 }30 31 void AddEdge(int from, int to, int cap, int cost) {32 edges.push_back(Edge(from, to, cap, 0, cost));33 edges.push_back(Edge(to, from, 0, 0, -cost));34 m = edges.size();35 G[from].push_back(m - 2);36 G[to].push_back(m - 1);37 }38 39 bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost) {40 for (int i = 0; i < n; i++) d[i] = INF;41 memset(inq, 0, sizeof(inq));42 d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;43 44 queue<int> Q;45 Q.push(s);46 while (!Q.empty()) {47 int u = Q.front(); Q.pop();48 inq[u] = 0;49 for (int i = 0; i < G[u].size(); i++) {50 Edge& e = edges[G[u][i]];51 if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {52 d[e.to] = d[u] + e.cost;53 p[e.to] = G[u][i];54 a[e.to] = min(a[u], e.cap - e.flow);55 if (!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }56 }57 }58 }59 if (d[t] == INF) return false;60 if (flow + a[t] > flow_limit) a[t] = flow_limit - flow;61 flow += a[t];62 cost += d[t] * a[t];63 for (int u = t; u != s; u = edges[p[u]].from) {64 edges[p[u]].flow += a[t];65 edges[p[u] ^ 1].flow -= a[t];66 }67 return true;68 }69 70 int MincostFlow(int s, int t, int flow_limit, int& cost) {71 int flow = 0; cost = 0;72 while (flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));73 return flow;74 }75 };76 77 MCMF g;78 79 int main() {80 int v, e;81 while (scanf("%d%d", &v, &e) == 2 && v) {82 g.init(v * 2 - 2);83 for (int i = 2; i <= v - 1; i++)84 g.AddEdge(i - 1, v + i - 2, 1, 0);85 int a, b, c;86 while (e--) {87 scanf("%d%d%d", &a, &b, &c);88 if (a != 1 && a != v) a += v - 2; else a--;89 b--;90 g.AddEdge(a, b, 1, c);91 }92 int cost;93 g.MincostFlow(0, v - 1, 2, cost);94 printf("%d\n", cost);95 }96 return 0;97 }
UVa1658 Admiral (拆点法,最小费用流)
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