Uva-12092-Paint the Roads

将城市进行拆点，源点跟城市连容量为k，费用为0，城市之间的路连容量为1，费用为路的费用。最后将拆点后的点与汇点相连，容量为k，费用为0.然后跑最小费用流就行了

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
const int inf=1<<29;
const int maxn=110;
const int maxm=maxn*maxn;
int e,st,des,n,m,k,head[maxn],nxt[maxm],flow[maxm],cost[maxm],pnt[maxm],dist[maxn],pre[maxn];
bool vis[maxn];
queue<int> q;
void AddEdge(int u,int v,int f,int c)
{
pnt[e]=v;nxt[e]=head[u];flow[e]=f;cost[e]=c;head[u]=e++;
pnt[e]=u;nxt[e]=head[v];flow[e]=0;cost[e]=-c;head[v]=e++;
}
bool Spfa()
{
for(int i=st;i<=des;i++)
{
pre[i]=-1;
dist[i]=inf;
}
dist[st]=0;
q.push(st);
while(!q.empty())
{
int u=q.front();
vis[u]=0;
q.pop();
for(int i=head[u];i!=-1;i=nxt[i])
if(flow[i]&&dist[pnt[i]]>dist[u]+cost[i])
{
dist[pnt[i]]=dist[u]+cost[i];
pre[pnt[i]]=i;
if(!vis[pnt[i]])
{
q.push(pnt[i]);
vis[pnt[i]]=1;
}
}
}
return dist[des]!=inf;
}
int maxflow()
{
int ans=0,f=0;
while(Spfa())
{
int mini=inf;
for(int i=pre[des];i!=-1;i=pre[pnt[i^1]])
mini=min(mini,flow[i]);
for(int i=pre[des];i!=-1;i=pre[pnt[i^1]])
{
flow[i]-=mini;
flow[i^1]+=mini;
}
ans+=dist[des]*mini;
f+=mini;
}
if(f!=k*n)
return -1;
return ans;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
st=e=0;des=2*n+1;
memset(head,-1,sizeof(head));
for(int i=1;i<=n;i++)
{
AddEdge(st,i,k,0);
AddEdge(n+i,des,k,0);
}
for(int i=0;i<m;i++)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
u++;
v++;
AddEdge(u,n+v,1,c);
}
printf("%d\n",maxflow());
}
return 0;
}