卧薪尝胆,厚积薄发。
两棵树的问题
Date: Mon Aug 27 14:29:21 CST 2018 In Category: NoCategory

Description:

有两棵树,找一个点的集合使得他们在两棵树中都联通,最大化点权和。
$1\le n \le 50$

Solution:

联通这个限制很不好处理,由于 $n$ 很小,可以枚举每个点作为根,那么如果要选一个点就必须选这个点的父亲,于是最大权闭合子图即可。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
int n;
#define MAXN 80
#define INF 0x3f3f3f3f
vector<int> v1[MAXN],v2[MAXN];
int val[MAXN];
int fa[MAXN];
bool v[MAXN];
void dfs1(int k)
{
v[k] = true;
for(vector<int>::iterator it = v1[k].begin();it != v1[k].end();++it)
{
if(v[*it])continue;
fa[*it] = k;
dfs1(*it);
}
return;
}
void dfs2(int k)
{
v[k] = true;
for(vector<int>::iterator it = v2[k].begin();it != v2[k].end();++it)
{
if(v[*it])continue;
fa[*it] = k;
dfs2(*it);
}
return;
}
struct edge
{
int to,nxt,f;
}e[MAXN << 3];
int edgenum = 0;
int lin[MAXN];
void add(int a,int b,int f)
{
e[edgenum].to = b;e[edgenum].f = f;e[edgenum].nxt = lin[a];lin[a] = edgenum;++edgenum;
e[edgenum].to = a;e[edgenum].f = 0;e[edgenum].nxt = lin[b];lin[b] = edgenum;++edgenum;
return;
}
int s,t;
int ch[MAXN];
int flow(int k,int f)
{
if(k == t)return f;
int r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f)
{
int l = flow(e[i].to,min(f - r,e[i].f));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r == 0)ch[k] = -1;
return r;
}
bool bfs()
{
memset(ch,-1,sizeof(ch));
ch[s] = 0;
queue<int> q;
q.push(s);
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return (ch[t] != -1);
}
int dinic()
{
int ans = 0,r;
while(bfs())while(r = flow(s,INF))ans += r;
return ans;
}
int main()
{
scanf("%d",&n);
for(int i = 1;i <= n;++i)scanf("%d",&val[i]);
int a,b;
for(int i = 1;i < n;++i)
{
scanf("%d%d",&a,&b);++a;++b;
v1[a].push_back(b);v1[b].push_back(a);
}
for(int i = 1;i < n;++i)
{
scanf("%d%d",&a,&b);++a;++b;
v2[a].push_back(b);v2[b].push_back(a);
}
int ans = 0;
for(int rt = 1;rt <= n;++rt)
{
int sum = 0;
memset(lin,-1,sizeof(lin));
edgenum = 0;
memset(v,false,sizeof(v));
dfs1(rt);
for(int i = 1;i <= n;++i)
{
if(i == rt)continue;
add(i,fa[i],INF);
}
memset(v,false,sizeof(v));
dfs2(rt);
for(int i = 1;i <= n;++i)
{
if(i == rt)continue;
add(i,fa[i],INF);
}
s = n + 1;t = s + 1;
for(int i = 1;i <= n;++i)
{
if(val[i] > 0)sum += val[i];
if(val[i] > 0)add(s,i,val[i]);
else add(i,t,-val[i]);
}
ans = max(ans,sum - dinic());
}
cout << ans << endl;
return 0;
}
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