思路：
源->1连费用0 流量2 其它的边 费用w 流量1 n->汇 费用0 流量2 最小费用流 搞定~//By SiriusRen#include#include #include #include using namespace std;#define N 2010#define M 201000int n,m,first[N],next[M],v[M],edge[M],cost[M],tot,xx,yy,zz,S,T;int vis[N],with[N],d[N],minn[N],ans;void Add(int x,int y,int C,int E){ edge[tot]=E,cost[tot]=C,v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y,int C,int E){Add(x,y,C,E),Add(y,x,-C,0);}bool tell(){ memset(vis,0,sizeof(vis)),memset(d,0x3f,sizeof(d)); memset(with,0,sizeof(with)),memset(minn,0x3f,sizeof(minn)); queue q;d[S]=0;q.push(S); while(!q.empty()){ int t=q.front();q.pop();vis[t]=0; for(int i=first[t];~i;i=next[i]) if(d[v[i]]>d[t]+cost[i]&&edge[i]>0){ d[v[i]]=d[t]+cost[i],minn[v[i]]=min(minn[t],edge[i]),with[v[i]]=i; if(!vis[v[i]])vis[v[i]]=1,q.push(v[i]); } }return d[T]!=0x3f3f3f3f;}int zeng(){ for(int i=T;i!=S;i=v[with[i]^1]) edge[with[i]]-=minn[T],edge[with[i]^1]+=minn[T]; return minn[T]*d[T];}int main(){ memset(first,-1,sizeof(first)); scanf("%d%d",&n,&m); S=n+1,T=n+2; for(int i=1;i<=m;i++){ scanf("%d%d%d",&xx,&yy,&zz); add(xx,yy,zz,1),add(yy,xx,zz,1); }add(S,1,0,2),add(n,T,0,2); while(tell())ans+=zeng(); printf("%d\n",ans);}
