Annoying problem
题目连接：
http://acm.hdu.edu.cn/showproblem.php?pid=5296
Description
Coco has a tree, whose nodes are conveniently labeled by 1,2,…,n, which has n-1 edge，each edge has a weight. An existing set S is initially empty.
Now there are two kinds of operation:
1 x： If the node x is not in the set S, add node x to the set S
2 x： If the node x is in the set S,delete node x from the set S
Now there is a annoying problem: In order to select a set of edges from tree after each operation which makes any two nodes in set S connected. What is the minimum of the sum of the selected edges’ weight ?
Input
one integer number T is described in the first line represents the group number of testcases.( T<=10 )
For each test:
The first line has 2 integer number n,q(0<n,q<=100000) describe the number of nodes and the number of operations.
The following n-1 lines each line has 3 integer number u,v,w describe that between node u and node v has an edge weight w.(1<=u,v<=n,1<=w<=100)
The following q lines each line has 2 integer number x,y describe one operation.（x=1 or 2,1<=y<=n）
Output
Each testcase outputs a line of "Case #x:" , x starts from 1.
The next q line represents the answer to each operation.
Sample Input
1
6 5
1 2 2
1 5 2
5 6 2
2 4 2
2 3 2
1 5
1 3
1 4
1 2
2 5
Sample Output
Case #1:
0
6
8
8
4
Hint
题意
给你一棵树，现在有两个操作
1.把某个点染成黑色
2.把某个点染成白色
然后每次操作结束后，问你黑色点构成的树的所有边权和是多少。
一开始全是白色的点。
题解：
首先黑色点构成的树一定是唯一的。
然后我们加入一个点，就只用考虑以前那堆点中dfs序比他小和比他大的点u,v。
如果找不到的话，找字典序最大和最小的点就好了。
然后这棵树增加的距离就是dis(x,u)+dis(x,v)+dis(u,v)
删除点也是一样的。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int flag[maxn],idx[maxn],cnt=1,n,q,LCA[maxn][21],deep[maxn],G[maxn][21];
vector<pair<int,int> >E[maxn];
set<pair<int,int> >S;
long long ans = 0;
void init()
{
ans = 0;
S.clear();cnt=1;
memset(idx,0,sizeof(idx));
memset(flag,0,sizeof(flag));
memset(LCA,0,sizeof(LCA));
memset(G,0,sizeof(G));
memset(deep,0,sizeof(deep));
for(int i=0;i<maxn;i++)E[i].clear();
}
void dfs(int x,int fa)
{
idx[x]=cnt++;
for(int i=0;i<E[x].size();i++)
{
int v = E[x][i].first;
if(v==fa)continue;
LCA[v][0]=x,deep[v]=deep[x]+1,G[v][0]=E[x][i].second;
dfs(v,x);
}
}
long long QueryDis(int u , int v){
long long ans = 0;
if(deep[u] < deep[v]) swap( u , v );
for(int i = 20 ; i >= 0 ; -- i ) if( deep[u] - (1 << i) >= deep[v] ) ans += G[u][i],u = LCA[u][i];
if( u == v ) return ans;
for(int i = 20 ; i >= 0 ; -- i ) if( LCA[u][i] != LCA[v][i] ) ans += (G[u][i] + G[v][i]) , u = LCA[u][i] , v = LCA[v][i];
return ans + G[u][0] + G[v][0];
}
void Lca_init()
{
for(int j = 1 ; j <= 20 ; ++ j)
for(int i = 1 ; i <= n ; ++ i)
if(LCA[i][j-1]){
LCA[i][j]=LCA[LCA[i][j-1]][j-1];
G[i][j] = G[i][j-1] + G[LCA[i][j-1]][j-1];
}
}
long long solve(int x)
{
if(S.size()==0)return 0;
set<pair<int,int> >::iterator it;
it=S.lower_bound(make_pair(idx[x],x));
int d1,d2;
if(it==S.begin()||it==S.end())
{
d1=S.begin()->second;
d2=S.rbegin()->second;
}
else
{
d1=it->second;
it--;
d2=it->second;
}
return QueryDis(d1,x)+QueryDis(d2,x)-QueryDis(d1,d2);
}
void solve()
{
init();
scanf("%d%d",&n,&q);
for(int i=1;i<n;i++)
{
int x,y,z;scanf("%d%d%d",&x,&y,&z);
E[x].push_back(make_pair(y,z));
E[y].push_back(make_pair(x,z));
}
dfs(1,-1);
Lca_init();
while(q--)
{
int op,x;scanf("%d%d",&op,&x);
if(op==1&&flag[x]==1);
else if(op==2&&flag[x]==0);
else if(op==1)flag[x]=1,ans+=solve(x),S.insert(make_pair(idx[x],x));
else if(op==2)flag[x]=0,S.erase(make_pair(idx[x],x)),ans-=solve(x);
printf("%lld\n",ans/2);
}
}
int main()
{
int t;scanf("%d",&t);
for(int i=1;i<=t;i++)
{
printf("Case #%d:\n",i);
solve();
}
return 0;
}