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uva10806
题目链接请戳 这里
解题思路
可以用最小费用最大流建模。
每条道路上的时间为费用,容量为1表示只能一个人通过(因为只对道路做了限值,顶点可以重复通过)
另建标号为0的顶点,其到标号为1的顶点容量为2,花费为0。
代码
#include<stdio.h> #include<string.h> #include<queue> using namespace std; const int maxn = 110; const int maxm = 5010; const int INF = 1000000000; //这里用的是lrj紫书里的模板 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<Edge> edges; vector<int> G[maxn]; int inq[maxn]; int d[maxn]; int p[maxn]; int a[maxn]; void init(int n) { this->n = n; for (int i = 0; i <= n; i++) G[i].clear(); edges.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) { for (int i = 0; i <= n; i++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = 1; p[s] = -1; a[s] = INF; queue<int> 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 from[maxm], to[maxm]; int w[maxm]; int main() { int n, m; MCMF A; scanf("%d", &n); while (n != 0) { scanf("%d", &m); long long cost = 0; A.init(n); //建立标号为0的顶点 A.AddEdge(0, 1, 2, 0); //因为为无向图,所以要建来回两条边 for (int i = 0; i < m; i++) { scanf("%d%d%d", &from[i], &to[i], &w[i]); A.AddEdge(from[i], to[i], 1, w[i]); A.AddEdge(to[i], from[i], 1, w[i]); } int f; f = A.MincostMaxflow(0, n, cost); if (f != 2) printf("Back to jail\n"); else printf("%lld\n", cost); scanf("%d", &n); } return 0; }
uva10806
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