http://www.lydsy.com/JudgeOnline/problem.php?id=1927
思路:把一个点拆成两个点，
S->i 费用0，流量1 (代表这个点可以移动到其他点所必备的流量)
i+n->T 费用0，流量1 (每个点都必须要走过)
u->v+n 费用w,流量1 (代表可以移动到那个点)
S->i+n 费用a[i],流量1 (代表从这个点瞬移)
1 #include2 #include 3 #include 4 #include 5 #include 6 int tot,go[200005],next[200005],first[200005],cost[200005],flow[200005]; 7 int op[200005],dis[200005],c[200005],vis[200005],edge[200005],from[200005]; 8 int S,T,n,m,ans; 9 int read(){10 char ch=getchar();int t=0,f=1;11 while (ch<'0'||ch>'9'){ if (ch=='-') f=-1;ch=getchar();}12 while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}13 return t*f;14 }15 void insert(int x,int y,int z,int l){16 tot++;17 go[tot]=y;18 next[tot]=first[x];19 first[x]=tot;20 flow[tot]=z;21 cost[tot]=l;22 }23 void add(int x,int y,int z,int l){24 insert(x,y,z,l);op[tot]=tot+1;25 insert(y,x,0,-l);op[tot]=tot-1;26 }27 bool spfa(){28 for (int i=S;i<=T;i++)29 dis[i]=0x3f3f3f3f,vis[i]=0;30 int h=1,t=1;c[1]=S;vis[S]=1;dis[S]=0;31 while (h<=t){32 int now=c[h++];33 for (int i=first[now];i;i=next[i]){34 int pur=go[i];35 if (dis[pur]>dis[now]+cost[i]&&flow[i]){36 dis[pur]=dis[now]+cost[i];37 edge[pur]=i;38 from[pur]=now;39 if (vis[pur]) continue;40 vis[pur]=1;41 c[++t]=pur;42 }43 }44 vis[now]=0;45 }46 return dis[T]!=0x3f3f3f3f;47 }48 void updata(){49 int mn=0x7ffffff;50 for (int i=T;i!=S;i=from[i]){51 mn=std::min(mn,flow[edge[i]]);52 }53 for (int i=T;i!=S;i=from[i]){54 ans+=mn*cost[edge[i]];55 flow[edge[i]]-=mn;56 flow[op[edge[i]]]+=mn;57 }58 }59 int main(){60 n=read();m=read();61 S=0;T=n+n+1;62 for (int i=1;i<=n;i++)63 add(S,i,1,0);64 for (int i=1;i<=n;i++){65 int x=read();66 add(S,i+n,1,x);67 }68 for (int i=1;i<=n;i++)69 add(i+n,T,1,0);70 while (m--){71 int u=read(),v=read(),w=read();72 if (u>v) std::swap(u,v);73 add(u,v+n,1,w);74 } 75 ans=0;76 while (spfa()) updata();77 printf("%d\n",ans);78 }