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　　拆点搞定度数限制，也就是每个点最多经过一次……源点汇点除外。

 1 /**************************************************************
 2     Problem: 1877
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:900 ms
 7     Memory:5972 kb
 8 ****************************************************************/
 9
10 //BZOJ 1877
11 #include<vector>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<iostream>
16 #include<algorithm>
17 #define rep(i,n) for(int i=0;i<n;++i)
18 #define F(i,j,n) for(int i=j;i<=n;++i)
19 #define D(i,j,n) for(int i=j;i>=n;--i)
20 #define pb push_back
21 using namespace std;
22 inline int getint(){
23     int v=0,sign=1; char ch=getchar();
24     while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();}
25     while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();}
26     return v*sign;
27 }
28 const int N=1010,M=200000,INF=~0u>>2;
29 typedef long long LL;
30 /******************tamplate*********************/
31 int n,m,ans,flow;
32 struct edge{int from,to,v,c;};
33 struct Net{
34     edge E[M];
35     int head[N],next[M],cnt;
36     void ins(int x,int y,int z,int c){
37         E[++cnt]=(edge){x,y,z,c};
38         next[cnt]=head[x]; head[x]=cnt;
39     }
40     void add(int x,int y,int z,int c){
41         ins(x,y,z,c); ins(y,x,0,-c);
42     }
43     int from[N],Q[M],d[N],S,T;
44     bool inq[N];
45     bool spfa(){
46         int l=0,r=-1;
47         F(i,1,T) d[i]=INF;
48         d[S]=0; Q[++r]=S; inq[S]=1;
49         while(l<=r){
50             int x=Q[l++];
51             inq[x]=0;
52             for(int i=head[x];i;i=next[i])
53                 if(E[i].v>0 && d[x]+E[i].c<d[E[i].to]){
54                     d[E[i].to]=d[x]+E[i].c;
55                     from[E[i].to]=i;
56                     if (!inq[E[i].to]){
57                         Q[++r]=E[i].to;
58                         inq[E[i].to]=1;
59                     }
60                 }
61         }
62         return d[T]!=INF;
63     }
64     void mcf(){
65         int x=INF;
66         for(int i=from[T];i;i=from[E[i].from])
67             x=min(x,E[i].v);
68         for(int i=from[T];i;i=from[E[i].from]){
69             E[i].v-=x;
70             E[i^1].v+=x;
71         }
72         flow+=x;
73         ans+=x*d[T];
74     }
75     void init(){
76         n=getint(); m=getint(); cnt=1;
77         S=1; T=n+n;
78         F(i,1,n) add(i,i+n,1,0);
79         add(1,1+n,INF,0);
80         add(n,n+n,INF,0);
81         int x,y,z;
82         F(i,1,m){
83             x=getint(); y=getint(); z=getint();
84             add(x+n,y,1,z);
85         }
86         while(spfa()) mcf();
87         printf("%d %d\n",flow,ans);
88     }
89 }G1;
90
91 int main(){
92 #ifndef ONLINE_JUDGE
93     freopen("1877.in","r",stdin);
94     freopen("1877.out","w",stdout);
95 #endif
96     G1.init();
97     return 0;
98 }
99