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 }
spfa
然后再讲讲费用流做法。因为不存在负的流量，但是我们知道$c_i‘ \geqslant 0$，因此我们考虑将$m_i$变成$-c_i + m_i‘$。
其中$m_i‘$始终非负。可以理解为这个做法就是将所有边的流量先变为0再重新调整。
对于绝对值的部分稍微处理一下就行了。（之前那个式子脑残取了等，调了半天，sad。。。）
Code
1 /**
2 * bzoj
3 * Problem#3597
4 * Accepted
5 * Time: 1168ms
6 * Memory: 2004k
7 */
8 #include <iostream>
9 #include <cstdlib>
10 #include <cstdio>
11 #include <queue>
12 #include <cmath>
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, r;
24 double c;
25
26 Edge(int ed = 0, int nx = 0, int r = 0, double c = 0) : ed(ed), nx(nx), r(r), c(c) { }
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, int r, double c) {
45 es.push_back(Edge(v, h[u], r, c));
46 h[u] = (signed) es.size() - 1;
47 }
48
49 void addArc(int u, int v, int cap, double c) {
50 addEdge(u, v, cap, c);
51 addEdge(v, u, 0, -c);
52 }
53
54 Edge& operator [] (int p) {
55 return es[p];
56 }
57 } MapManager;
58
59 const signed int inf = (signed) (~0u >> 1);
60 const double dinf = 1e10;
61 const double eps = 1e-4;
62
63 template <typename T>
64 T __abs(T x) {
65 return (x < 0) ? (-x) : (x);
66 }
67
68 typedef class Network {
69 public:
70 int S, T;
71 MapManager g;
72
73 int *le;
74 int *mf;
75 double *f;
76 boolean *vis;
77
78 Network() { }
79 // be sure T is the last node
80 Network(int S, int T) : S(S), T(T), g(T) {
81 f = new double[(T + 1)];
82 le = new int[(T + 1)];
83 mf = new int[(T + 1)];
84 vis = new boolean[(T + 1)];
85 pfill(vis, vis + T, false);
86 }
87 ~Network() {
88 delete[] f;
89 delete[] le;
90 delete[] mf;
91 delete[] vis;
92 }
93
94 double spfa() {
95 double w;
96 static queue<int> que;
97 pfill(f, f + T + 1, dinf);
98 que.push(S);
99 f[S] = 0, le[S] = -1, mf[S] = inf;
100 while (!que.empty()) {
101 int e = que.front();
102 que.pop();
103 vis[e] = false;
104 for (int i = g.h[e], eu; ~i; i = g[i].nx) {
105 if (!g[i].r)
106 continue;
107 eu = g[i].ed, w = f[e] + g[i].c;
108 if (w < f[eu]) {
109 f[eu] = w, le[eu] = i, mf[eu] = min(mf[e], g[i].r);
110 if (!vis[eu]) {
111 vis[eu] = true;
112 que.push(eu);
113 }
114 }
115 }
116 }
117 if (__abs(f[T] - dinf) < eps)
118 return dinf;
119 double rt = 0;
120 for (int p = T, e; ~le[p]; p = g[e ^ 1].ed) {
121 e = le[p];
122 g[e].r -= mf[T];
123 g[e ^ 1].r += mf[T];
124 rt += mf[T] * g[e].c;
125 }
126 return rt;
127 }
128
129 double min_cost() {
130 double rt = 0, delta;
131 while (__abs((delta = spfa()) - dinf) >= eps) {
132 rt += delta;
133 }
134 return rt;
135 }
136 } Network;
137
138 // S: n + 1, T: n + 2
139 int n, m;
140 int *u, *v, *a, *b, *c, *d;
141 double init_ans = 0.0;
142
143 inline void init() {
144 scanf("%d%d", &n, &m);
145 u = new int[(m + 1)];
146 v = new int[(m + 1)];
147 a = new int[(m + 1)];
148 b = new int[(m + 1)];
149 c = new int[(m + 1)];
150 d = new int[(m + 1)];
151 for (int i = 1; i <= m; i++) {
152 scanf("%d%d%d%d%d%d", u + i, v + i, a + i, b + i, c + i, d + i);
153 init_ans += c[i] * d[i];
154 }
155 }
156
157 boolean check(double mid) {
158 double cost = 0;
159 Network network(n + 1, n + 2);
160 MapManager& g = network.g;
161 for (int i = 1; i <= m; i++) {
162 if (u[i] == n + 1) {
163 g.addArc(u[i], v[i], c[i], 0);
164 } else {
165 g.addArc(u[i], v[i], c[i], -(a[i] + mid) + d[i]);
166 g.addArc(u[i], v[i], inf, b[i] + mid + d[i]);
167 // cost += c[i] * (-d[i] + a[i] + mid);
168 cost += c[i] * (-d[i] + a[i] + mid);
169 }
170 }
171 cost += network.min_cost();
172 return cost < 0;
173 }
174
175 inline void solve() {
176 double l = 0, r = init_ans, mid;
177 for (int t = 0; t < 128 && l < r - eps; t++) {
178 mid = (l + r) / 2;
179 if (check(mid))
180 l = mid;
181 else
182 r = mid;
183 }
184 printf("%.2lf\n", l);
185 }
186
187 int main() {
188 init();
189 solve();
190 return 0;
191 }
原文地址：https://www.cnblogs.com/yyf0309/p/10197979.html
时间： 2024-11-06 22:10:29