跑一个流量为2的最小费用流，顺利1A
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> P;
struct Edge {
int to, cap, cost, rev;
Edge(int a, int b, int c, int d) : to(a), cap(b), cost(c), rev(d) {}
};
int V;
vector<Edge> G[maxn];
int h[maxn];
int dist[maxn];
int prevv[maxn], preve[maxn];
void init() {
for (int i = 0; i < maxn; i++) G[i].clear();
}
void addedge(int u, int v, int cap, int cost) {
G[u].push_back(Edge(v, cap, cost, G[v].size()));
G[v].push_back(Edge(u, 0, -cost, G[u].size()-1));
}
int min_cost_flow(int s, int t, int f) {
int res = 0;
memset(h, 0, sizeof(h));
while (f > 0) {
priority_queue<P, vector<P>, greater<P> > q;
memset(dist, INF, sizeof(dist));
dist[s] = 0;
q.push(P(0, s));
while (!q.empty()) {
P p = q.top(); q.pop();
int v = p.second;
if (dist[v] < p.first) continue;
for (int i = 0; i < G[v].size(); i++) {
Edge &e = G[v][i];
if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
prevv[e.to] = v;
preve[e.to] = i;
q.push(P(dist[e.to], e.to));
}
}
}
if (dist[t] == INF) return -1;
for (int v = 0; v < V; v++) h[v] += dist[v];
int d = f;
for (int v = t; v != s; v = prevv[v]) {
d = min(d, G[prevv[v]][preve[v]].cap);
}
f -= d;
res += d*h[t];
for (int v = t; v != s; v = prevv[v]) {
Edge &e = G[prevv[v]][preve[v]];
e.cap -= d;
G[v][e.rev].cap += d;
}
}
return res;
}
int main() {
int n, m, u, v, cost;
while (~scanf("%d %d", &n, &m)) {
init();
int s = 0, t = n - 1;
V = n;
while (m--) {
scanf("%d %d %d", &u, &v, &cost);
addedge(u-1, v-1, 1, cost);
addedge(v-1, u-1, 1, cost);
}
printf("%d\n", min_cost_flow(s, t, 2));
}
return 0;
}
