UVALive 2197 Paint the Roads(費用流)
阿新 • • 發佈:2018-12-17
題意:n個點,m條邊,還有一個k,給出每條邊的資訊,讓你選擇一些邊,使得這些邊組成的圖裡面,每個點只屬於k個迴路上,問最小的費用
如果每個點都只在k個迴路上,那麼這些點的出度入度是相等的,都是等於k,所以對於每個點,拆點分別連源點匯點,容量為k,費用為0,然後邊就正常的連上,容量為1,費用為邊權,跑一遍費用流,若最後的流量等於k*n,就是答案,否則就是-1
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int maxn=110; const int maxm=1e5+7; const int inf=0x3f3f3f3f; struct Node { int to; int capa; int cost; int next; }edge[maxm]; int cnt; int source,sink; int n,m,k; int head[maxn]; int dis[maxn]; int rec[maxn]; int pre[maxn]; bool vis[maxn]; void init() { cnt=0; memset(head,-1,sizeof(head)); return; } void add(int u,int v,int capa,int cost) { edge[cnt].to=v; edge[cnt].capa=capa; edge[cnt].cost=cost; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].to=u; edge[cnt].capa=0; edge[cnt].cost=-cost; edge[cnt].next=head[v]; head[v]=cnt++; return; } bool spfa() { memset(dis,inf,sizeof(dis)); memset(pre,-1,sizeof(pre)); memset(rec,-1,sizeof(rec)); memset(vis,false,sizeof(vis)); queue<int> que; que.push(source); dis[source]=0; vis[source]=true; while(!que.empty()) { int node=que.front(); que.pop(); vis[node]=false; for(int i=head[node];~i;i=edge[i].next) { int v=edge[i].to; if(edge[i].capa>0&&dis[v]>dis[node]+edge[i].cost) { dis[v]=dis[node]+edge[i].cost; rec[v]=i; pre[v]=node; if(!vis[v]) { vis[v]=true; que.push(v); } } } } return dis[sink]!=inf; } int mcmf() { int maxflow=0; int mincost=0; while(spfa()) { int node=sink; int minn=inf; while(node!=source) { minn=min(edge[rec[node]].capa,minn); node=pre[node]; } maxflow+=minn; node=sink; while(node!=source) { mincost+=minn*edge[rec[node]].cost; edge[rec[node]].capa-=minn; edge[rec[node]^1].capa+=minn; node=pre[node]; } } if(maxflow==k*n) return mincost; return -1; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int test; scanf("%d",&test); while(test--) { init(); scanf("%d%d%d",&n,&m,&k); source=0; sink=n*2+1; for(int i=1;i<=n;i++) { add(source,i,k,0); add(i+n,sink,k,0); } for(int i=0;i<m;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u+1,v+n+1,1,w); } printf("%d\n",mcmf()); } return 0; }