给一个图， N个点， m条边， 每条边有权值， 从1走到n， 然后从n走到1， 一条路不能走两次，求最短路径。
如果(u, v)之间有边， 那么加边（u, v, 1, val), (v, u, 1, val)， val是路的长度，代表费用， 1是流量。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {, }, {, }, {, -}, {, } };
const int maxn = 4e5+;
int num, head[maxn*], s, t, n, m, nn, dis[maxn], flow, cost, cnt, cap[maxn], q[maxn], cur[maxn], vis[maxn];
struct node
{
int to, nextt, c, w;
node(){}
node(int to, int nextt, int c, int w):to(to), nextt(nextt), c(c), w(w) {}
}e[maxn*];
int spfa() {
int st, ed;
st = ed = ;
mem2(dis);
++cnt;
dis[s] = ;
cap[s] = inf;
cur[s] = -;
q[ed++] = s;
while(st<ed) {
int u = q[st++];
vis[u] = cnt-;
for(int i = head[u]; ~i; i = e[i].nextt) {
int v = e[i].to, c = e[i].c, w = e[i].w;
if(c && dis[v]>dis[u]+w) {
dis[v] = dis[u]+w;
cap[v] = min(c, cap[u]);
cur[v] = i;
if(vis[v] != cnt) {
vis[v] = cnt;
q[ed++] = v;
}
}
}
}
if(dis[t] == inf)
return ;
cost += dis[t]*cap[t];
flow += cap[t];
for(int i = cur[t]; ~i; i = cur[e[i^].to]) {
e[i].c -= cap[t];
e[i^].c += cap[t];
}
return ;
}
int mcmf() {
flow = cost = ;
while(spfa())
;
return cost;
}
void add(int u, int v, int c, int val) {
e[num] = node(v, head[u], c, val); head[u] = num++;
e[num] = node(u, head[v], , -val); head[v] = num++;
}
void init() {
mem1(head);
num = cnt = ;
mem(vis);
}
void input() {
int x, y, w;
while(m--) {
scanf("%d%d%d", &x, &y, &w);
add(x, y, , w);
add(y, x, , w);
}
add(s, , , );
add(n, t, , );
}
int main()
{
while(~scanf("%d%d", &n, &m)) {
init();
s = , t = n+;
input();
int ans = mcmf();
printf("%d\n", ans);
}
return ;
}