题目链接 \(Click\) \(Here\)
非常有趣的一个题目。
关键结论：所有的单位费用应该被分配在流量最大的边上。
即：在保证最大流的前提下，使最大流量最小。这里我们采用二分的方法，每次判断让所有边的流量\(<=mid\)时是否依然有最大流，求得最小的最大流量\(*p\)即可。
为什么会有实数流量呢？其实我也不懂，不过这也造成这个题目需要把流量改成\(double\)，有很多细节需要小心谨慎。。。
#include <bits/stdc++.h>
using namespace std;
const int N = 400010;
const int M = 400010;
const int INF = 0x3f3f3f3f;
int u[N], v[N], flow[N]; double f[N];
int n, m, p, cnt = -1, head[N];
struct edge {
int nxt, to; double f;
}e[M];
void add_edge (int from, int to, double flw) {
e[++cnt].nxt = head[from];
e[cnt].to = to;
e[cnt].f = flw;
head[from] = cnt;
}
void add_len (int u, int v, double f) {
add_edge (u, v, f);
add_edge (v, u, 0);
}
queue <int> q;
int cur[N], deep[N];
bool bfs (int s, int t) {
memcpy (cur, head, sizeof (head));
memset (deep, 0x3f, sizeof (deep));
deep[s] = 0; q.push (s);
while (!q.empty ()) {
int u = q.front (); q.pop ();
for (int i = head[u]; ~i; i = e[i].nxt) {
int v = e[i].to;
if (deep[v] == INF && fabs (e[i].f) > 1e-8) {
deep[v] = deep[u] + 1;
q.push (v);
}
}
}
return deep[t] != INF;
}
double dfs (int u, int t, double lim) {
if (u == t || fabs (lim) < 1e-8) {
return lim;
}
double tmp = 0, flow = 0;
for (int &i = cur[u]; ~i; i = e[i].nxt) {
int v = e[i].to;
if (deep[v] == deep[u] + 1) {
tmp = dfs (v, t, min (lim, e[i].f));
lim -= tmp;
flow += tmp;
e[i ^ 0].f -= tmp;
e[i ^ 1].f += tmp;
if (fabs (lim) < 1e-8) break;
}
}
return flow;
}
double Dinic (int s, int t) {
double res = 0;
while (bfs (s, t)) {
res += dfs (s, t, INF);
}
return res;
}
double max_flow;
bool can_use (double flw) {
cnt = -1; int s = 1, t = n;
memset (head, -1, sizeof (head));
for (int i = 1; i <= m; ++i) {
add_len (u[i], v[i], min (f[i], flw));
}
return fabs (Dinic (s, t) - max_flow) < 1e-8;
}
int main () {
memset (head, -1, sizeof (head));
cin >> n >> m >> p;
for (int i = 1; i <= m; ++i) {
cin >> u[i] >> v[i] >> f[i];
add_len (u[i], v[i], f[i]);
}
int s = 1, t = n; max_flow = Dinic (s, t);
printf ("%.0lf\n", max_flow);
double l = 0, r = INF;
while (r - l > 1e-8) {
double mid = (l + r) / 2.0;
if (can_use (mid)) {
r = mid;
} else {
l = mid;
}
}
printf ("%.4lf\n", r * p);
}