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uva10746
题目链接请戳 这里
解题思路
带权二分图的最优匹配
注意精度问题
代码
#include<stdio.h> #include<string.h> #include<queue> using namespace std; const int maxn = 110; const int maxm = 5010; const double INF = 1e16; struct Edge { int from, to, cap, flow; double cost; Edge(int u, int v, int c, int f, double w):from(u), to(v), cap(c), flow(f), cost(w) {} }; //这里用的是lrj紫书里的模板 struct MCMF { int n, m; vector<Edge> edges; vector<int> G[maxn]; int inq[maxn]; double 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, double 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, double &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] = 1000000000; 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 += d[t] * 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, double &cost) { int flow = 0; cost = 0; while (BellmanFord(s, t, flow, cost)); return flow; } }; int from[maxm], to[maxm]; long long w[maxm]; int main() { int n, m; MCMF A; scanf("%d%d", &n, &m); while (!(n == 0 && m == 0)) { A.init(n + m + 1); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { double weight; scanf("%lf", &weight); A.AddEdge(n + j, i, 1, weight); } double cost = 0; //超级源点到各个警察有一条容量为1的边 for (int i = n + 1; i <= n + m; i++) A.AddEdge(0, i, 1, 0); //各个银行到超级汇点有一条容量为1的边 for (int i = 1; i <= n; i++) A.AddEdge(i, n + m + 1, 1, 0); int f = A.MincostMaxflow(0, n + m + 1, cost); //可以加一个很小的数来避免精度问题 printf("%.2lf\n", cost / n + 1e-9); scanf("%d%d", &n, &m); } return 0; }
uva10746
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