题解:
最小费用流;
拆点法;
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=1000;
const int inf=100000000;
int n,m;
int totn,s,t;
struct Edge{
int from,to,cap,flow,cost;
};
vector<int>G[maxn];
vector<Edge>edges;
int addedge(int x,int y,int z,int w){
Edge e;
e.from=x;e.to=y;e.cap=z;e.flow=0;e.cost=w;
edges.push_back(e);
e.to=x;e.from=y;e.cap=0;e.flow=0;e.cost=-w;
edges.push_back(e);
int c=edges.size();
G[x].push_back(c-2);
G[y].push_back(c-1);
}
queue<int>q;
int inq[maxn];
int d[maxn];
int p[maxn];
int spfa(int &nowflow,int &nowcost){
for(int i=1;i<=totn;++i){
d[i]=inf;inq[i]=0;
}
d[s]=0;q.push(s);inq[s]=1;p[s]=0;
while(!q.empty()){
int x=q.front();q.pop();inq[x]=0;
for(int i=0;i<G[x].size();++i){
Edge e=edges[G[x][i]];
if((e.cap>e.flow)&&(d[x]+e.cost<d[e.to])){
d[e.to]=d[x]+e.cost;
p[e.to]=G[x][i];
if(!inq[e.to]){
q.push(e.to);
inq[e.to]=1;
}
}
}
}
if(d[t]==inf)return 0;
int f=inf,x=t;
while(x!=s){
Edge e=edges[p[x]];
f=min(f,e.cap-e.flow);
x=e.from;
}
nowflow+=f;nowcost+=f*d[t];
x=t;
while(x!=s){
edges[p[x]].flow+=f;
edges[p[x]^1].flow-=f;
x=edges[p[x]].from;
}
return 1;
}
int Mcmf(){
int flow=0,cost=0;
while(spfa(flow,cost));
printf("%d %d\n",flow,cost);
}
int minit(){
for(int i=1;i<=n+n;++i)G[i].clear();
edges.clear();
while(!q.empty())q.pop();
}
int main(){
scanf("%d%d",&n,&m);
minit();
while(m--){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
addedge(x+n,y,1,z);
}
addedge(1,1+n,inf,0);
addedge(n,n+n,inf,0);
for(int i=2;i<=n-1;++i)addedge(i,i+n,1,0);
totn=n+n;s=1;t=n+n;
Mcmf();
return 0;
}