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经典的最大权闭合子图问题;
这种问题的求解思路是:
建图:将s连边向权值为正的点,通过依赖关系连接权值为正的点和权值为负的点,权值为负的点连边向t;
求c=最小割,a=所有权值为正的节点权值和,ans=a-c;
证明过程网上是有的;
用dinic敲的,但是用的还不熟练;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<cstdlib> 6 #include<ctime> 7 #include<algorithm> 8 using namespace std; 9 const int maxn=200; 10 const int inf=1000000000; 11 const int s=0,t=120; 12 struct node{ 13 int y,next,flow,re; 14 }e[maxn*10]; 15 int linkk[maxn],len=0,n,m,g[maxn][maxn],cnt[maxn],w[maxn],sum=0; 16 void print(int x){printf("%d\n",x);} 17 void print(int x,int y){printf("%d %d\n",x,y);} 18 void insert(int x,int y,int flow){ 19 e[++len].y=y;e[len].flow=flow; 20 e[len].next=linkk[x];linkk[x]=len;e[len].re=len+1; 21 e[++len].y=x;e[len].flow=0; 22 e[len].next=linkk[y];linkk[y]=len;e[len].re=len-1; 23 } 24 bool flag=0; 25 int read(){ 26 int x=0;char ch=getchar(); 27 while(ch<'0'||ch>'9')ch=getchar(); 28 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 29 if(ch=='\n')flag=1; 30 return x; 31 } 32 void init(){ 33 scanf("%d%d",&n,&m); 34 int v; 35 for(int i=1;i<=n;i++){ 36 scanf("%d",&v);insert(s,i,v); 37 flag=0;sum+=v; 38 while(!flag){cnt[i]++;g[i][cnt[i]]=read();} 39 } 40 for(int i=1;i<=m;i++)scanf("%d",&w[i]); 41 for(int i=1;i<=n;i++) 42 for(int j=1;j<=cnt[i];j++) 43 insert(i,g[i][j]+n,inf); 44 for(int i=1;i<=m;i++)insert(i+n,t,w[i]); 45 } 46 int q[maxn],tail=0,level[maxn],head=0; 47 bool bfs(){ 48 memset(level,-1,sizeof(level)); 49 head=0;tail=0;level[s]=1;q[++tail]=s; 50 while(++head<=tail){ 51 int x=q[head]; 52 for(int i=linkk[x];i;i=e[i].next){ 53 if(level[e[i].y]==-1&&e[i].flow){ 54 q[++tail]=e[i].y; 55 level[e[i].y]=level[x]+1; 56 } 57 } 58 } 59 return level[t]>0; 60 } 61 int find(int x,int flow){ 62 if(x==t)return flow; 63 int maxflow=0,d=0; 64 for(int i=linkk[x];i&&maxflow<flow;i=e[i].next){ 65 if(level[e[i].y]==level[x]+1&&e[i].flow){ 66 if(d=find(e[i].y,min(flow-maxflow,e[i].flow))){ 67 maxflow+=d; 68 e[i].flow-=d; 69 e[e[i].re].flow+=d; 70 } 71 } 72 } 73 if(!maxflow)level[x]=-1; 74 return maxflow; 75 } 76 void work(){ 77 int d=0,ans=0; 78 while(bfs()) 79 while(d=find(s,inf)) 80 ans+=d; 81 cout<<sum-ans<<endl; 82 } 83 int main(){ 84 //freopen("1.in","r",stdin); 85 //freopen("1.out","w",stdout); 86 init(); 87 work(); 88 }