正题

题目链接:http://www.51nod.com/Challenge/Problem.html#problemId=1325

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题目大意

两棵树。要求选出一组权值和最大的点使得这两个点集在两棵树上都是连通的。

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解题思路

我们枚举一个中间点$i$，然后从这个点开始扩展连通块。那么如果这个点作为根的话，选择一个点之前就有选择它的父节点这个一个要求，然后用最大权闭合图求解即可。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=55,inf=2147483647/3;
struct node{int to,next,w;
};
int n,val[N],ans,s,t;
queue<int> q;
struct Net_Flow{node a[N*200];int tot,ls[N],dep[N];void init(){memset(ls,0,sizeof(ls));tot=1;}void addl(int x,int y,int w){a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;a[tot].w=w;a[++tot].to=x;a[tot].next=ls[y];ls[y]=tot;a[tot].w=0;return;}bool bfs(){memset(dep,0,sizeof(dep));dep[s]=1;while(!q.empty())q.pop();q.push(s);while(!q.empty()){int x=q.front();q.pop();for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(dep[y]||!a[i].w)continue;dep[y]=dep[x]+1;if(y==t)return 1;q.push(y);}}return 0;}int dinic(int x,int flow){if(x==t)return flow;int rest=0,k;for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(dep[x]+1!=dep[y]||!a[i].w)continue;rest+=(k=dinic(y,min(a[i].w,flow-rest)));a[i].w-=k;a[i^1].w+=k;if(rest==flow)return flow;}if(!rest)dep[x]=0;return rest;}int get_ans(){int ans=0;while(bfs())ans+=dinic(s,inf);return ans;}
}nf;
struct Graph{node a[N*2];int tot,ls[N];void addl(int x,int y){a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;return;}void dfs(int x,int fa){for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(y==fa)continue;nf.addl(y,x,inf);dfs(y,x);}}
}G1,G2;
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&val[i]); for(int i=1;i<n;i++){int x,y;scanf("%d%d",&x,&y);x++;y++;G1.addl(x,y);G1.addl(y,x);}for(int i=1;i<n;i++){int x,y;scanf("%d%d",&x,&y);x++;y++;G2.addl(x,y);G2.addl(y,x);}s=n+1;t=s+1;for(int i=1;i<=n;i++){nf.init();G1.dfs(i,i);G2.dfs(i,i);int sum=0;for(int i=1;i<=n;i++)if(val[i]>0)nf.addl(s,i,val[i]),sum+=val[i];else nf.addl(i,t,-val[i]);sum-=nf.get_ans();ans=max(ans,sum);}printf("%d",ans);
}