对于边u->v的处理方法是先建立边u->u+n,权值为0, 然后建立边u+n->v权值为c,这样可以避免一个点经过多次的情况
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2 * 1e3 + 10;
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], d[maxn], p[maxn], a[maxn];
void init(int x) {
n = x;
edges.clear();
for (int i = 0; i <= n; i++) G[i].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, long long &cost) {
memset(inq, 0, sizeof(inq));
for (int i = 0; i <= n; i++) d[i] = INF;
inq[s] = 1, d[s] = 0, 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;
flow += a[t];
cost += (long long)d[t] * (long long)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, long long &cost) {
int flow = 0; cost = 0;
while (BellmanFord(s, t, flow, cost));
return flow;
}
};
int main() {
MCMF mcmf;
int n, m;
while (~scanf("%d %d", &n, &m)) {
mcmf.init(n<<1);
for (int i = 2; i <= n - 1; i++) mcmf.addedge(i, i + n, 1, 0);
mcmf.addedge(1, n + 1, 2, 0);
mcmf.addedge(n, n<<1, 2, 0);
for (int i = 0; i < m; i++) {
int u, v, cost; scanf("%d %d %d", &u, &v, &cost);
mcmf.addedge(u + n, v, 1, cost);
}
long long cost = 0;
mcmf.mincostmaxflow(1, n<<1, cost);
cout << cost << endl;
}
return 0;
}