Annoying problem

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 203    Accepted Submission(s): 60

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
 

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
typedef unsigned long long ll; const int N = 233333;
int n, m, head[N], mm;
struct Edge
{
int to, next, w;
} e[N << 1];
void add(int u, int v, int w = 1)
{
e[mm].to = v;
e[mm].next = head[u];
e[mm].w = w;
head[u] = mm++;
}
int sz[N], dep[N];
int f[N][22]; /// f[i][j] 表示 i 的第 2^j 个祖先
int dfn[N]; ///dfs index
int cur;
int id[N]; /// you dfs xu qiu chu bian hao
int len[N];
void dfs(int u, int fa) /// 点从 1 開始标号
{
f[u][0] = fa;
sz[u] = 1;
dfn[u] = ++cur;
id[cur] = u;
for (int i=head[u]; ~i; i=e[i].next)
{
int v = e[i].to;
int w = e[i].w;
if (v != fa)
{
dep[v] = dep[u] + 1;
len[v] = len[u] + w;
dfs(v, u);
sz[u] += sz[v];
}
}
}
int maxh;
void gao()
{
cur = 0;
dep[0] = -1;
len[1] = dep[1] = 0;
f[1][0] = 1;
dfs(1, 0);f[1][0] = 1;
int j;
for (j=1; (1<<j)<n; j++)
for (int i=1; i<=n; i++)
f[i][j] = f[f[i][j-1]][j-1];
maxh = j - 1;
}
int swim(int x, int k)
{
for (int i=0; i<=maxh; i++)
if (k >> i & 1)
x = f[x][i];
return x;
}
int LCA(int x, int y)
{
if (dep[x] > dep[y]) swap(x, y); ///dep[x] <= dep[y];
y = swim(y, dep[y] - dep[x]);
if (x == y) return y;
for (int i=maxh; i>=0; i--)
{
if (f[x][i] != f[y][i])
x = f[x][i], y = f[y][i];
}
return f[x][0];
}
int Q; set<int> se;
set<int>::iterator it;
int dist(int x, int y)
{
int lca = LCA(x, y);
return len[x] - len[lca] + len[y] - len[lca];
}
int solve(int u)
{
if (se.empty()) return 0;
int x, y;
int t = *se.begin();
it = se.lower_bound( u);
y = *it;
it--;
x = *(it );
int t2 = *se.rbegin();
x = id[x];
y = id[y];
if (t2 < u || t > u)
{
x = id[t]; y = id[t2];
}
u = id[u];
return len[u] - len[LCA(x,u) ] - len[LCA(y,u)] + len[LCA(x,y) ];
}
int main()
{
int re;
cin>>re;
int ca=1;
while (re--)
{
cin>>n>>Q;
mm = 0;
memset(head,-1,sizeof head);
for (int i=0; i<n-1; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
gao();
printf("Case #%d:\n", ca++);
se.clear();
int ans = 0;
while (Q--)
{
int op, u;
scanf("%d%d", &op, &u);
u = dfn[u];
if (op==1)
{
it = se.find(u);
if (it==se.end())
{
ans += solve(u);
se.insert(u);
}
}
else
{
it = se.find(u);
if (it != se.end())
{
se.erase(u);
ans -= solve(u);
}
}
printf("%d\n", ans);
}
}
return 0;
}

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