dfs一遍得到每一个节点的dfs序,对于要插入的节点x分两种情况考虑:
1,假设x能够在集合中的某些点之间,找到左边和右边距离x近期的两个点,即DFS序小于x的DFS序最大点,和大于x的DFS序最小的点......
2.假设x在集合中的点某一側,则找距离x的dfs序最小和最大的点
将x插入这个集合最少要走的距离为 dist[x]-dist[LCA(left,x)]-dist[LCA(right,x)]+dist[LCA(left,right)]
删除同理
Annoying problem
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 334 Accepted Submission(s): 95
Problem 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
Author
FZUACM
Source
2015 Multi-University Training Contest 1
/* ***********************************************
Author :CKboss
Created Time :2015年07月21日 星期二 21时06分11秒
File Name :HDOJ5296.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
const int maxn=120100;
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cost;
}edge[maxn*2];
int Adj[maxn],Size;
void init() { memset(Adj,-1,sizeof(Adj)); Size=0; }
void Add_Edge(int u,int v,int c)
{
edge[Size].to=v; edge[Size].cost=c;
edge[Size].next=Adj[u];
Adj[u]=Size++;
}
int n,q;
int dist[maxn],ti[maxn],cnt;
void dfs(int len,int u,int fa)
{
dist[u]=len; ti[u]=cnt;
for(int i=Adj[u];~i;i=edge[i].next)
{
int to=edge[i].to;
int cost=edge[i].cost;
if(to==fa) continue;
cnt++;
dfs(len+cost,to,u);
}
}
/********************** LCA **********************************/
const int DEG=20;
int fa[maxn][DEG];
int deg[maxn];
void BFS(int root)
{
queue<int> q;
memset(deg,0,sizeof(deg));
memset(fa,0,sizeof(fa));
deg[root]=0;
fa[root][0]=root;
q.push(root);
while(!q.empty())
{
int u=q.front(); q.pop();
for(int i=1;i<DEG;i++)
{
fa[u][i]=fa[fa[u][i-1]][i-1];
}
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(v==fa[u][0]) continue;
deg[v]=deg[u]+1;
fa[v][0]=u;
q.push(v);
}
}
}
int LCA(int u,int v)
{
if(deg[u]>deg[v]) swap(u,v);
int hu=deg[u],hv=deg[v];
int tu=u,tv=v;
for(int det=hv-hu,i=0;det;i++,det=det/2)
{
if(det&1) tv=fa[tv][i];
}
if(tu==tv) return tu;
for(int i=DEG-1;i>=0;i--)
{
if(fa[tu][i]==fa[tv][i]) continue;
tu=fa[tu][i]; tv=fa[tv][i];
}
return fa[tu][0];
}
struct Node
{
int val,cnt;
bool operator<(const Node& nd) const { return cnt<nd.cnt; }
bool operator==(const Node& nd) const { return val==nd.val; }
bool operator!=(const Node& nd) const { return val!=nd.val; }
};
set<Node> st;
int CL(int flag,Node ND)
{
if(flag==0) st.erase(ND);
set<Node>::iterator it1,it2;
int x=ND.val;
it2=st.upper_bound(ND);
it1=it2; it1--;
/// check if in mid
if(it1->val!=0&&it2->val!=n+10) /// in mid
{
int left=it1->val;
int right=it2->val;
if(flag) st.insert(ND);
return dist[x]-dist[LCA(x,left)]-dist[LCA(x,right)]+dist[LCA(left,right)];
}
else // in side
{
if(it2->val==n+10) /// all in left
{
it2=st.begin(); it2++;
int left=it2->val;
int right=it1->val;
if(flag) st.insert(ND);
return dist[x]-dist[LCA(x,left)]-dist[LCA(x,right)]+dist[LCA(left,right)];
}
else if(it1->val==0) /// all in right
{
int left=it2->val;
it1=st.end();
it1--; it1--;
int right=it1->val;
if(flag) st.insert(ND);
return dist[x]-dist[LCA(x,left)]-dist[LCA(x,right)]+dist[LCA(left,right)];
}
}
}
/// return change val of solve
int solve(int kind,int x)
{
Node ND = (Node){x,ti[x]};
/// if in mid find nearst point
set<Node>::iterator it1,it2;
if(kind==1) // add
{
if(st.count(ND)==1) return 0;
if(st.size()==2)
{
st.insert(ND);
return 0;
}
else if(st.size()==3)
{
it1=st.begin(); it1++;
int v=it1->val;
st.insert(ND);
return dist[x]+dist[v]-2*dist[LCA(v,x)];
}
else
{
return CL(1,ND);
}
}
else if(kind==2) // remove
{
if(st.count(ND)==0) return 0;
if(st.size()==3)
{
st.erase(ND);
return 0;
}
else if(st.size()==4)
{
it1=st.begin();
it1++;
int v=it1->val;
it1++;
int u=it1->val;
st.erase(ND);
return dist[u]+dist[v]-2*dist[LCA(u,v)];
}
else
{
return CL(0,ND);
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T_T,cas=1;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&q);
init();
for(int i=0,u,v,c;i<n-1;i++)
{
scanf("%d%d%d",&u,&v,&c);
Add_Edge(u,v,c); Add_Edge(v,u,c);
}
cnt=1; st.clear();
st.insert((Node){0,-INF});
st.insert((Node){n+10,INF});
dfs(0,1,1); BFS(1);
int all=0;
printf("Case #%d:\n",cas++);
while(q--)
{
int k,x;
scanf("%d%d",&k,&x);
if(k==1) all+=solve(k,x);
else if(k==2) all-=solve(k,x);
printf("%d\n",all);
}
}
return 0;
}