分析：把每个点分为 n 和 n' ，两点之间连一条容量为1，费用为0的边，这样就能保证一个点只能被经过一次。
#include #include #include #include #include #include #include #include #include #define FRER() freopen("in.txt","r",stdin) #define FREW() freopen("out.txt","w",stdout) #define go int T;cin>>T;for(int kase=1;kase<=T;kase++) #define debug cout<<"****************"< pii; template T gcd(T a,T b){return !b?a:gcd(b,a%b);} const int N = 10007 , M = 100007 , inf = 0x3f3f3f3f; int n,m,nEdge,s,t; int head[N],to[M],from[M],flow[M],cap[M],cost[M],nxt[M],pre[M],a[N],d[N],inq[N]; void addEdge(int u,int v,int c,int co){ from[nEdge] = u ; to[nEdge] = v ; cap[nEdge] = c ; cost[nEdge] = co; flow[nEdge] = 0 ; nxt[nEdge] = head[u] ; head[u] = nEdge++; from[nEdge] = v ; to[nEdge] = u ; cap[nEdge] = 0 ; cost[nEdge] = -co; flow[nEdge] = 0 ; nxt[nEdge] = head[v] ; head[v] = nEdge++; } ll MFMC(){ ll f = 0 , res = 0; while(1){ memset(d, inf, sizeof(d)); memset(inq, 0, sizeof(inq)); memset(a, 0, sizeof(a)); queueq;q.push(s); d[s] = 0 ; inq[s] = 1; a[s] = inf; while(!q.empty()){ int u = q.front();q.pop(); inq[u] = 0; for(int e = head[u] ; ~e ; e = nxt[e]){ int v = to[e]; if(cap[e] > flow[e] && d[v] > d[u] + cost[e]){ d[v] = d[u] + cost[e]; pre[v] = e; a[v] = min(a[u],cap[e] - flow[e]); if(!inq[v]){ q.push(v); inq[v] = 1; } } } } if(!a[t]) break; f += (ll)a[t]; res += (ll)d[t]*(ll)a[t]; for(int u = t ; u != s ; u = from[pre[u]]){ flow[pre[u]] += a[n]; flow[pre[u]^1] -= a[n]; } } return res; } int main(){ //FRER(); //FREW(); while(cin >> n >> m){ nEdge = 0; memset(head, -1, sizeof(head)); for(int i = 2 ; i < n ; i++) addEdge(i, i+n, 1, 0); for(int i = 1 ; i <= m ; i++){ int u,v,c; cin >> u >> v >> c; if(u == 1 || u == n) addEdge(u, v, 1, c); else addEdge(u+n, v, 1, c); } s = 0 ; t = 2*n + 10; addEdge(s, 1, 2, 0); addEdge(n, t, 2, 0); cout << MFMC() << endl; } }
