H-Minimum-cost Flow
设 cost(u/v, 1) 表示容量为u/v,流量为1的总花费。
我们可以把网络扩大v/u倍,之后 cost(u/v,1) * (v/u) = cost(1, v/u)。
这样所有的边的容量都是1,我们可以跑mcmf,记录每一条增广路的价值。因为容量是1,因此每有一条增广路,总流量会加一。
因为每条增广路的价值是不减的,因此我们一定是顺序取,直到流量等于v/u。
别忘了最后乘以u/v,把网络缩小到原来的样子。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 0x3f3f3f;
const int maxn = 105;
vector<ll> val;
ll sum[maxn];
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]; //是否在队列中
int d[maxn]; // bellmanford 到源点距离
int p[maxn]; //上一条弧
int a[maxn]; //可改进量
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, ll& cost) {
for (int i = 0; i <= n; i++) d[i] = INF;
for (int i = 0; i <= n; i++) inq[i] = 0;
d[s] = 0;
inq[s] = 1;
p[s] = 0;
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; // 当没有可增广的路时退出
flow += a[t];
cost += (ll)d[t] * (ll)a[t];
val.push_back((ll)d[t] * (ll)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, ll& cost) {
int flow = 0;
cost = 0;
while (BellmanFord(s, t, flow, cost))
;
return flow;
}
} mcmf;
int main() {
int n, m, q;
while (~scanf("%d%d", &n, &m)) {
mcmf.init(n);
val.clear();
for (int i = 0, u, v, w; i < m; i++) {
scanf("%d%d%d", &u, &v, &w);
mcmf.AddEdge(u, v, 1, w);
}
ll uesless;
mcmf.MincostMaxflow(1, n, uesless);
for (int i = 1; i <= val.size(); i++) sum[i] = sum[i - 1] + val[i - 1];
scanf("%d", &q);
for (int i = 0, u, v; i < q; i++) {
scanf("%d%d", &u, &v);
if (!u || (v + u - 1) / u > val.size()) {
puts("NaN");
continue;
}
int x = v / u, y = v % u;
ll num = sum[x] * u + val[x] * y, den = v;
ll d = __gcd(num, den);
printf("%lld/%lld\n", num / d, den / d);
}
}
}