HihoCoder 119周 网络流五·最大权闭合子图
传送门:http://hihocoder.com/contest/hiho119/problem/1
题解
最大权闭合子图建图,跑一边网络流
代码
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//while(true) RP++; #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #define LL long long using namespace std; struct Dinic{ const static int Maxn=500; const static int Maxm=Maxn*Maxn; const static int Inf=2100000000; int head[Maxn+5]; int cur[Maxn+5]; struct Edge{ int v,nxt,f; Edge(){} Edge(int v0,int n0,int f0){ v=v0; nxt=n0; f=f0; } }; Edge e[Maxm*2+5]; int nume; inline void init(){ nume=1; memset(head,0,sizeof(head)); } inline void insertEdge(int u,int v,int f){ e[++nume]=Edge(v,head[u],f); head[u]=nume; } inline void addEdge(int u,int v,int f){ insertEdge(u,v,f); insertEdge(v,u,0); } int dist[Maxn+5]; queue<int> que; int src,sink; inline bool bfs(){ while(!que.empty())que.pop(); memset(dist,-1,sizeof(dist)); dist[src]=0; que.push(src); while(!que.empty()){ int now=que.front(); que.pop(); for (int i=head[now];i;i=e[i].nxt){ if (e[i].f>0 && dist[e[i].v]==-1){ dist[e[i].v]=dist[now]+1; que.push(e[i].v); } } } return dist[sink]!=-1; } int dfs(int x,int delta){ if (x==sink){ return delta; }else{ int ret=0; for (int &i=cur[x];i;i=e[i].nxt){ if (e[i].f>0 && dist[e[i].v]==dist[x]+1){ int ddelta=dfs(e[i].v,min(e[i].f,delta)); e[i].f-=ddelta; e[i^1].f+=ddelta; delta-=ddelta; ret+=ddelta; } } return ret; } } inline int Max_Flow(int src0,int sink0){ src=src0; sink=sink0; int ret=0; while(bfs()){ memcpy(cur,head,sizeof(head)); ret+=dfs(src,Inf); } return ret; } }; Dinic dinic; int n,m; int main(){ scanf("%d%d",&n,&m); int src=n+m+1,sink=n+m+2; dinic.init(); for (int i=1;i<=m;i++){ int cst; scanf("%d",&cst); dinic.addEdge(i,sink,cst); } int sum=0; for (int i=1;i<=n;i++){ int a,k; scanf("%d%d",&a,&k); sum+=a; dinic.addEdge(src,m+i,a); for (int j=1;j<=k;j++){ int st; scanf("%d",&st); dinic.addEdge(m+i,st,dinic.Inf); } } printf("%d\n",sum-dinic.Max_Flow(src,sink)); return 0; } |

原文链接:HihoCoder 119周 网络流五·最大权闭合子图
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