HDU 5458(Stability-LCT维护连通图2点间割边个数)

Stability

给定一张无向连通图，维护2个操作，删一条边（删后保证连通），求2点间有多少条边，只要删除2点就不连通。

Input

There are multiple test cases(no more than 3 cases), and the first line contains an integer t, meaning the totally number of test cases.

For each test case, the first line contains three integers n, m and q, where 1≤n≤3×104,1≤m≤105 and 1≤q≤105. The nodes in graph G are labelled from 1 to n.

Each of the following m lines contains two integers u and v describing an undirected edge between node u and node v.

Following q lines - each line describes an operation or a query in the formats:

⋅ 1 a b: delete one edge between a and b. We guarantee the existence of such edge.

⋅ 2 a b: query the stability between a and b.

Output

For each test case, you should print first the identifier of the test case.

Then for each query, print one line containing the stability between corresponding pair of nodes.

Sample Input

1

10 12 14

1 2

1 3

2 4

2 5

3 6

4 7

4 8

5 8

6 10

7 9

8 9

8 10

2 7 9

2 7 10

2 10 6

2 10 5

1 10 6

2 10 1

2 10 6

2 3 10

1 8 5

2 5 10

2 4 5

1 7 9

2 7 9

2 10 5

Sample Output

Case #1:

0

0

0

0

2

4

3

3

2

3

4

Source

2015 ACM/ICPC Asia Regional Shenyang Online

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倒过来做，原题变为加边。

将原图缩点成树，问题转化为求两点间路径长度。

我们把没条边设为代价1，答案即为路径和。

现在考虑在树中加一条边，等价于把两端点间链缩成一点，

故可把路径上的边设成0即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<set>
#include<map>
#include<unordered_map>
#include<cctype>
#include<ctime>
#include<iomanip>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MAXN (60000+10)
#pragma comment(linker, "/STACK:1024000000,1024000000")

#define MAXM (200000+10)
#define MAXQ (200000+10)
#define pb push_back
#define mp make_pair
typedef int ll;

void mul(ll &a,ll b){a=a*b;}
void add(ll &a,ll b){a=a+b;}
class LCT
{
public:
int father[MAXN],siz[MAXN];
int ch[MAXN][2];
bool root[MAXN];
bool rev[MAXN];
ll mulv[MAXN],sumv[MAXN],val[MAXN];
void mem(int n)
{
MEM(father) MEM(siz) MEM(root)
For(i,n+1) siz[i]=root[i]=mulv[i]=val[i]=sumv[i]=1;    root[0]=1;
MEM(ch)  MEM(rev)
}
void Set(int x,ll bi){
val[x]=sumv[x]=bi;
}

void pushdown(int x)
{
if (!x) return ;
if (rev[x]) {
if (ch[x][0]) rev[ch[x][0]]^=1;
if (ch[x][1]) rev[ch[x][1]]^=1;
swap(ch[x][0],ch[x][1]);
rev[x]^=1;
}
if (mulv[x]!=1) {
if (ch[x][0]) mul(mulv[ch[x][0]],mulv[x]),mul(val[ch[x][0]],mulv[x]),mul(sumv[ch[x][0]],mulv[x]);
if (ch[x][1]) mul(mulv[ch[x][1]],mulv[x]),mul(val[ch[x][1]],mulv[x]),mul(sumv[ch[x][1]],mulv[x]);
mulv[x]=1;
}
}
void maintain(int x)
{
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
sumv[x]=(sumv[ch[x][0]]+sumv[ch[x][1]]+val[x]);
}
void rotate(int x)
{
int y=father[x],kind=ch[y][1]==x;

ch[y][kind]=ch[x][!kind];
if (ch[y][kind]) {
father[ch[y][kind]]=y;
}
father[x]=father[y];
father[y]=x;
ch[x][!kind]=y;
if (root[y])
{
root[x]=1;root[y]=0;
}
else
{
ch[father[x]][ ch[father[x]][1]==y ] = x;
}
maintain(y);maintain(x);
}

void P(int x)
{
if (!root[x]) P(father[x]);
pushdown(x);
}

void splay(int x)
{
P(x);
while(!root[x])
{
int y=father[x];
int z=father[y];
if (root[y]) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
}

int access(int x)
{
int y=0;
do
{
splay(x);
if (ch[x][1]) root[ch[x][1]]=1;
ch[x][1]=y;
if (y) root[y]=0;
maintain(x);
y = x;
x=father [x];
} while (x) ;
return y;
}

void cut(int x)
{
access(x);
splay(x);

father[ch[x][0]]=0;
root[ch[x][0]]=1;
ch[x][0]=0;
maintain(x);
}

void join(int x,int y)
{
make_root(x);
access(y);
splay(y);
ch[y][1]=x;
father[x]=y;
maintain(y);
root[x]=0;
}
void reverse(int x){
rev[x]^=1;   // 标记记完后迅速处理
}
void make_root(int x){
access(x);splay(x);
reverse(x);pushdown(x);
}
int get_root(int x){
access(x);
splay(x);
do {
pushdown(x);
if (ch[x][0]) x=ch[x][0];
else break;
}while(1);
return x;
}

void Mul(int x,ll cost){
pushdown(x);
mulv[x]=cost;mul(val[x],cost);mul(sumv[x],cost);
}

bool check(int x,int y) {
while (father[x]) x=father[x];
while (father[y]) y=father[y];
return x==y;

}

}S;

int n,m,q;
int u[MAXM],v[MAXM];
struct ask{
int p,x,y;
}comm[MAXQ];
ll ans[MAXQ];

map< pair<int,int> ,int > S2;
map< pair<int,int> ,int >::iterator it;
int main()
{
int T;
cin>>T;
For(kcase,T) {
S2.clear();
scanf("%d%d%d",&n,&m,&q);
printf("Case #%d:\n",kcase);

For(i,m) {
scanf("%d%d",&u[i],&v[i]);
if (u[i]>v[i]) swap(u[i],v[i]);
}

int qm=0;
For(i,q) {
scanf("%d%d%d",&comm[i].p,&comm[i].x,&comm[i].y);
if (comm[i].x>comm[i].y) swap(comm[i].x,comm[i].y);
if (comm[i].p==1) {
++qm;
if ((it=S2.find(mp(comm[i].x,comm[i].y)))!=S2.end() ) (it->second)++;
else S2[ mp(comm[i].x,comm[i].y) ]=1;
}
}

S.mem(2*n-1);
For(i,n) S.Mul(i,0);

int nownode=n;
For(i,m) {
if ((it=S2.find(mp(u[i],v[i])) ) !=S2.end())
if ( (it->second) >0 ){
(it->second) --;
continue;
}
if (S.check(u[i],v[i])) {
S.make_root(u[i]);
S.access(v[i]);
S.splay(v[i]);
S.Mul(v[i],0);
} else {
S.join(++nownode,u[i]);
S.join(nownode,v[i]);
}
}

ForD(i,q) {
int x=comm[i].x,y=comm[i].y;
if (comm[i].p==1) {
if (S.check(x,y)) {
S.make_root(x);
S.access(y);
S.splay(y);
S.Mul(y,0);
} else {
S.join(++nownode,x);
S.join(nownode,y);
}

} else {
S.make_root(x);
S.access(y);
S.splay(y);
ans[i]=S.sumv[y];

}
}

For(i,q) {
if (comm[i].p==2) printf("%d\n",ans[i]);
}

}

return 0;
}