【模板】最小费用最大流(网络流)/洛谷P3381
题目链接
https://www.luogu.com.cn/problem/P3381
题目大意
输入格式
第一行包含四个正整数 \(n,m,s,t\),分别表示点的个数、有向边的个数、源点序号、汇点序号。
接下来\(m\)行,每行包含四个正整数 \(u_i,v_i,w_i,f_i\),表示第 \(i\) 条有向边从 \(u_i\) 出发,到达 \(v_i\),边权为 \(w_i\)(即该边最大流量为 \(w_i\) ),单位流量的费用为 \(f_i\) 。
输出格式
一行,包含两个整数,依次为最大流量和在最大流量情况下的最小费用。
题目解析
(待补充,咕咕咕。。。)
参考代码
\(SPFA\)的两个优化可以有效提升速度。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = (1LL) << 32;
const int N = 5005;
struct Edge{
int u, v;
ll cap, cost, flow;
};
vector <Edge> e;
vector <int> G[N];
ll a[N], d[N];
int inQ[N], p[N];
int n, m, s, t, cnt;
void addEdge(int u, int v, ll cap, ll cost, int i)
{
e.push_back((Edge){u, v, cap, cost, 0});
e.push_back((Edge){v, u, 0, -cost, 0});
G[u].push_back(i);
G[v].push_back(i^1);
}
bool SPFA(ll &flow, ll &cost)
{
for (int i = 0; i <= cnt; ++i) d[i] = INF;
memset(a, 0, sizeof a);
a[s] = INF, d[s] = 0;
deque <int> Q;
Q.push_back(s);
inQ[s] = 1;
while (!Q.empty())
{
int x = Q.front();
Q.pop_front();
inQ[x] = 0;
for (int i = 0; i <G[x].size(); ++i)
{
Edge &b = e[G[x][i]];
if (b.cap > b.flow && d[b.v] > d[x] + b.cost)
{
d[b.v] = d[x] + b.cost;
p[b.v] = G[x][i];
a[b.v] = min(a[x], b.cap-b.flow);
if (!inQ[b.v] && b.v != t)//优化1:终点无需入队
{
inQ[b.v] = 1;
if (!Q.empty() && d[b.v] < d[Q.front()]) Q.push_front(b.v);//优化2:small label first
else Q.push_back(b.v);
}
}
}
}
if (d[t] == INF) return false;
flow += a[t];
cost += a[t]*d[t];
for (int u = t; u != s; u = e[p[u]].u)
{
e[p[u]].flow += a[t];
e[p[u]^1].flow -= a[t];
}
return true;
}
ll mincostMaxflow(ll &cost)
{
cnt = n; //cnt: count nodes
ll flow = 0;
while (SPFA(flow, cost));
return flow;
}
int main()
{
int u, v;
ll w, c;
scanf("%d%d%d%d", &n, &m, &s, &t);
for (int i = 0; i < m; ++i)
{
scanf("%d%d%lld%lld", &u, &v, &w, &c);
addEdge(u, v, w, c, i << 1);
}
w = mincostMaxflow(c=0);
printf("%lld %lld\n", w, c);
return 0;
}
感谢支持!