题目传送门
题目大意
给定一个费用流，每条边有一个初始流量$c_i$和单位流量费用$d_i$，增加一条边的1单位的流量需要花费$b_i$的代价而减少一条边的1单位的流量需要花费$a_i$的代价。要求最小化总费用减少量和调整次数的比值（至少调整一次）。
根据基本套路，二分答案，移项，可以得到每条边的贡献。
设第$i$条边的流量变化量为$m_i$，每次变化花费的平均费用为$w_i$。那么有
$\sum c_id_i - \sum (c_i + m_i)d_i + |m_i|(w_i + mid) > 0$
$\sum m_id_i+ |m_i|(w_i + mid) < 0$
那么二分答案后等于判断是否存在一个增广环的费用和为负。（请手动参见式子脑补边权）。
然后可以写个spfa。
Code
1 /** 2 * bzoj 3 * Problem#3597 4 * Accepted 5 * Time: 464ms 6 * Memory: 1720k 7 */ 8 #include <iostream> 9 #include <cstdlib> 10 #include <cstdio> 11 #include <vector> 12 #include <queue> 13 using namespace std; 14 typedef bool boolean; 15 16 template <typename T> 17 void pfill(T* pst, const T* ped, T val) { 18 for ( ; pst != ped; *(pst++) = val); 19 } 20 21 typedef class Edge { 22 public: 23 int ed, nx; 24 double w; 25 26 Edge(int ed, int nx, double w) : ed(ed), nx(nx), w(w) { } 27 } Edge; 28 29 typedef class MapManager { 30 public: 31 int* h; 32 vector<Edge> es; 33 34 MapManager() { } 35 MapManager(int n) { 36 h = new int[(n + 1)]; 37 pfill(h, h + n + 1, -1); 38 } 39 ~MapManager() { 40 delete[] h; 41 es.clear(); 42 } 43 44 void addEdge(int u, int v, double w) { 45 es.push_back(Edge(v, h[u], w)); 46 h[u] = (signed) es.size() - 1; 47 } 48 49 Edge& operator [] (int p) { 50 return es[p]; 51 } 52 } MapManager; 53 54 const double dinf = 1e10; 55 const double eps = 1e-4; 56 57 typedef class Graph { 58 public: 59 int n, s; 60 MapManager g; 61 62 int *cnt; 63 double *f; 64 boolean *vis; 65 66 Graph(int n, int s) : n(n), s(s), g(n) { 67 cnt = new int[(n + 1)]; 68 f = new double[(n + 1)]; 69 vis = new boolean[(n + 1)]; 70 pfill(vis, vis + n + 1, false); 71 } 72 73 boolean neg_circle() { 74 static queue<int> que; 75 pfill(f, f + n + 1, dinf); 76 pfill(cnt, cnt + n + 1, 0); 77 f[s] = 0, cnt[s]++; 78 que.push(s); 79 while (!que.empty()) { 80 int e = que.front(); 81 que.pop(); 82 vis[e] = false; 83 for (int i = g.h[e], eu; ~i; i = g[i].nx) { 84 eu = g[i].ed; 85 double w = f[e] + g[i].w; 86 if (w < f[eu]) { 87 f[eu] = w, cnt[eu]++; 88 if (cnt[eu] > n) 89 return true; 90 if (!vis[eu]) { 91 que.push(eu); 92 vis[eu] = true; 93 } 94 } 95 } 96 } 97 return false; 98 } 99 } Graph; 100 101 int n, m; 102 int *u, *v, *a, *b, *c, *d; 103 double init_ans = 0.0; 104 105 inline void init() { 106 scanf("%d%d", &n, &m); 107 u = new int[(m + 1)]; 108 v = new int[(m + 1)]; 109 a = new int[(m + 1)]; 110 b = new int[(m + 1)]; 111 c = new int[(m + 1)]; 112 d = new int[(m + 1)]; 113 for (int i = 1; i <= m; i++) { 114 scanf("%d%d%d%d%d%d", u + i, v + i, a + i, b + i, c + i, d + i); 115 init_ans += c[i] * d[i]; 116 } 117 } 118 119 boolean check(double mid) { 120 Graph graph(n + 2, n + 1); 121 MapManager& g = graph.g; 122 123 for (int i = 1; i <= m; i++) { 124 if (u[i] == n + 1) { 125 g.addEdge(u[i], v[i], 0); 126 } else if (c[i]) { 127 g.addEdge(u[i], v[i], d[i] + b[i] + mid); 128 g.addEdge(v[i], u[i], -d[i] + a[i] + mid); 129 } else { 130 g.addEdge(u[i], v[i], d[i] + b[i] + mid); 131 } 132 } 133 134 return graph.neg_circle(); 135 } 136 137 inline void solve() { 138 double l = 0, r = init_ans, mid; 139 for (int t = 0; t < 128 && l < r - eps; t++) { 140 mid = (l + r) / 2; 141 if (check(mid)) 142 l = mid; 143 else 144 r = mid; 145 } 146 printf("%.2lf\n", l); 147 } 148 149 int main() { 150 init(); 151 solve(); 152 return 0; 153 }