P4014 分配问题 网络流
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 305, inf = 0x3f3f3f3f; 4 struct Edge { 5 int from, to, cap, flow, cost; 6 }; 7 8 struct MCMF { 9 int n, m, s, t; 10 vector<Edge> edges; 11 vector<int> G[maxn]; 12 int inq[maxn]; 13 int d[maxn]; 14 int p[maxn]; 15 int a[maxn]; 16 17 void init(int n) { 18 this->n = n; 19 for (int i = 1; i <= n; ++i) G[i].clear(); 20 edges.clear(); 21 } 22 23 void AddEdge(int from, int to, int cap, int cost) { 24 edges.push_back((Edge){from, to, cap, 0, cost}); 25 edges.push_back((Edge){to, from, 0, 0, -cost}); 26 m = edges.size(); 27 G[from].push_back(m-2); 28 G[to].push_back(m-1); 29 } 30 bool BellmanFord(int s, int t, int& flow, int& cost) { 31 for (int i = 1; i <= n; ++i) d[i] = inf; 32 memset(inq, 0, sizeof(inq)); 33 d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = inf; 34 35 queue<int> que; 36 que.push(s); 37 while (!que.empty()) { 38 int u = que.front(); que.pop(); 39 inq[u] = 0; 40 for (int i = 0; i < G[u].size(); ++i) { 41 Edge& e = edges[G[u][i]]; 42 if (e.cap > e.flow && d[e.to] > d[u] + e.cost) { 43 d[e.to] = d[u] + e.cost; 44 p[e.to] = G[u][i]; 45 a[e.to] = min(a[u], e.cap-e.flow); 46 if (!inq[e.to]) { que.push(e.to); inq[e.to] = 1; } 47 } 48 } 49 } 50 if (d[t] == inf) return false; 51 flow += a[t]; 52 cost += d[t] * a[t]; 53 int u = t; 54 while (u != s) { 55 edges[p[u]].flow += a[t]; 56 edges[p[u]^1].flow -= a[t]; 57 u = edges[p[u]].from; 58 } 59 return true; 60 } 61 int mincost(int s, int t) { 62 int flow = 0, cost = 0; 63 while (BellmanFord(s, t, flow, cost)); 64 return cost; 65 } 66 }mcmf; 67 int c[maxn][maxn]; 68 int main() { 69 int n; scanf("%d",&n); 70 int s = 2*n+1, t = 2*n+2; 71 72 for (int i = 1; i <= n; ++i) { 73 for (int j = 1; j <= n; ++j) { 74 scanf("%d",&c[i][j]); 75 } 76 } 77 78 /// 最小总效益 79 mcmf.init(2*n+2); 80 for (int i = 1; i <= n; ++i) { 81 for (int j = 1; j <= n; ++j) { 82 mcmf.AddEdge(i,j+n,1,c[i][j]); 83 } 84 } 85 for (int i = 1; i <= n; ++i) { 86 mcmf.AddEdge(s,i,1,0); 87 mcmf.AddEdge(i+n,t,1,0); 88 } 89 int ans = mcmf.mincost(s,t); 90 printf("%d\n",ans); 91 92 /// 最大总效益 93 mcmf.init(2*n+2); 94 for (int i = 1; i <= n; ++i) { 95 for (int j = 1; j <= n; ++j) { 96 mcmf.AddEdge(i,j+n,1,-c[i][j]); 97 } 98 } 99 for (int i = 1; i <= n; ++i) { 100 mcmf.AddEdge(s,i,1,0); 101 mcmf.AddEdge(i+n,t,1,0); 102 } 103 ans = mcmf.mincost(s,t); 104 printf("%d\n",-ans); 105 return 0; 106 }