看到要求两个量就下意识的想到了费用流= =，先把一个点拆成两个点就能够解决一个的只经过一次的限制

CODE：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 410
#define maxm 50000
#define inf 0x7fffffff
struct edges{
 int to,next,cap,dist;
}edge[maxm];
int next[maxn],l;
int s,t;
int addedge(int x,int y,int z,int cap){
 l++;
 edge[l*2]=(edges){y,next[x],cap,z};
 edge[l*2+1]=(edges){x,next[y],0,-z};
 next[x]=l*2;next[y]=l*2+1;
 return 0;
}
int dist[maxn],way[maxn],cost;
bool b[maxn];
queue<int> q;
bool spfa(){
 for (int i=1;i<=t;i++) dist[i]=inf;
 dist[s]=0;
 q.push(s);
 while (!q.empty()){
  int u=q.front();q.pop();
  b[u]=0;
  for (int i=next[u];i;i=edge[i].next)
   if (edge[i].cap&&dist[edge[i].to]>dist[u]+edge[i].dist){
    dist[edge[i].to]=dist[u]+edge[i].dist;
    way[edge[i].to]=i;
    if(!b[edge[i].to]){
     b[edge[i].to]=1;
     q.push(edge[i].to);
    }
   }
 }
 if (dist[t]==inf) return 0;
 return 1;
}
int mcmf(){
 int fl=0;
 while (spfa()){
  int flow=inf,x=t;
  while (x!=s){
   flow=min(flow,edge[way[x]].cap);
   x=edge[way[x]^1].to;
  }
  cost+=dist[t]*flow;fl+=flow;
  x=t;
  while (x!=s){
   edge[way[x]].cap-=flow;
   edge[way[x]^1].cap+=flow;
   x=edge[way[x]^1].to;
  }
 }
 return fl;
}
int main(){
 int n,m,S;
 scanf("%d%d",&n,&m);
 for (int i=2;i<n;i++) addedge(i*2,i*2+1,0,1);
 for (int j=1;j<=m;j++) {
  int x,y,z;
  scanf("%d%d%d",&x,&y,&z);
  addedge(x*2+1,y*2,z,1);
 }
 s=3;t=n*2;
 printf("%d ",mcmf());
 printf("%d",cost);
 return 0;
}