HIT2739 The Chinese Postman Problem（最小费用最大流）

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #include<algorithm>
  5 using namespace std;
  6 #define INF (1<<30)
  7 #define MAXN 111
  8 #define MAXM 8888
  9 struct Edge{
 10     int u,v,cap,cost,next;
 11 }edge[MAXM];
 12 int vs,vt,NV,NE,head[MAXN];
 13 void addEdge(int u,int v,int cap,int cost){
 14     edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
 15     edge[NE].next=head[u]; head[u]=NE++;
 16     edge[NE].u=v; edge[NE].v=u; edge[NE].cap=0; edge[NE].cost=-cost;
 17     edge[NE].next=head[v]; head[v]=NE++;
 18 }
 19 int d[MAXN],pre[MAXN];
 20 bool vis[MAXN];
 21 bool SPFA(){
 22     for(int i=0; i<NV; ++i){
 23         d[i]=INF; vis[i]=0;
 24     }
 25     d[vs]=0; vis[vs]=1;
 26     queue<int> que;
 27     que.push(vs);
 28     while(!que.empty()){
 29         int u=que.front(); que.pop();
 30         for(int i=head[u]; i!=-1; i=edge[i].next){
 31             int v=edge[i].v;
 32             if(edge[i].cap && d[v]>d[u]+edge[i].cost){
 33                 d[v]=d[u]+edge[i].cost;
 34                 pre[v]=i;
 35                 if(!vis[v]){
 36                     vis[v]=1;
 37                     que.push(v);
 38                 }
 39             }
 40         }
 41         vis[u]=0;
 42     }
 43     return d[vt]!=INF;
 44 }
 45 int MCMF(int &mxflow){
 46     int res=0;
 47     while(SPFA()){
 48         int flow=INF,cost=0;
 49         for(int u=vt; u!=vs; u=edge[pre[u]].u){
 50             flow=min(flow,edge[pre[u]].cap);
 51         }
 52         mxflow-=flow;
 53         for(int u=vt; u!=vs; u=edge[pre[u]].u){
 54             edge[pre[u]].cap-=flow;
 55             edge[pre[u]^1].cap+=flow;
 56             cost+=flow*edge[pre[u]].cost;
 57         }
 58         res+=cost;
 59     }
 60     return res;
 61 }
 62 int par[MAXN];
 63 int Find(int x){
 64     while(x!=par[x]){
 65         par[x]=par[par[x]];
 66         x=par[x];
 67     }
 68     return x;
 69 }
 70 bool Union(int a,int b){
 71     int pa=Find(a),pb=Find(b);
 72     if(pa==pb) return 0;
 73     par[pb]=pa;
 74     return 1;
 75 }
 76 int deg[MAXN];
 77 int main(){
 78     int t,n,m,a,b,c;
 79     scanf("%d",&t);
 80     while(t--){
 81         scanf("%d%d",&n,&m);
 82         vs=n; vt=vs+1; NV=vt+1; NE=0;
 83         memset(head,-1,sizeof(head));
 84         memset(deg,0,sizeof(deg));
 85         for(int i=0; i<n; ++i) par[i]=i;
 86         int res=0,tot=n,mxflow=0;
 87         for(int i=0; i<m; ++i){
 88             scanf("%d%d%d",&a,&b,&c);
 89             addEdge(a,b,INF,c);
 90             res+=c;
 91             ++deg[a]; --deg[b];
 92             if(Union(a,b)) --tot;
 93         }
 94         if(tot!=1){
 95             puts("-1");
 96             continue;
 97         }
 98         for(int i=0; i<n; ++i){
 99             if(deg[i]<0) addEdge(vs,i,-deg[i],0);
100             else addEdge(i,vt,deg[i],0),mxflow+=deg[i];
101         }
102         res+=MCMF(mxflow);
103         if(mxflow!=0) puts("-1");
104         else printf("%d\n",res);
105     }
106     return 0;
107 }