这题我一开始想的是缩点，然后a子树去掉有b的子树后假设有x个点，b同理有y个点，x*y就是答案，但是有点难写，我记得去年这场cf打的时候我口胡过一个做法，但现在不太记得去了，就去翻了一下，发现我当时的想法真的好，再看我刚才想的算法，惭愧惭愧
这就是我去年的做法，真的妙多了，删掉和ab相连的边，然后用并查集处理下联通块就好了
There are n cities in Berland and some pairs of them are connected by two-way roads. It is guaranteed that you can pass from any city to any other, moving along the roads. Cities are numerated from 1 to n
.
Two fairs are currently taking place in Berland — they are held in two different cities a
and b (1≤a,b≤n; a≠b
).
Find the number of pairs of cities x
and y (x≠a,x≠b,y≠a,y≠b) such that if you go from x to y you will have to go through both fairs (the order of visits doesn't matter). Formally, you need to find the number of pairs of cities x,y such that any path from x to y goes through a and b
(in any order).
Print the required number of pairs. The order of two cities in a pair does not matter, that is, the pairs (x,y)
and (y,x)
must be taken into account only once.
Input
The first line of the input contains an integer t
(1≤t≤4⋅104) — the number of test cases in the input. Next, t
test cases are specified.
The first line of each test case contains four integers n
, m, a and b (4≤n≤2⋅105, n−1≤m≤5⋅105, 1≤a,b≤n, a≠b
) — numbers of cities and roads in Berland and numbers of two cities where fairs are held, respectively.
The following m
lines contain descriptions of roads between cities. Each of road description contains a pair of integers ui,vi (1≤ui,vi≤n, ui≠vi
) — numbers of cities connected by the road.
Each road is bi-directional and connects two different cities. It is guaranteed that from any city you can pass to any other by roads. There can be more than one road between a pair of cities.
The sum of the values of n
for all sets of input data in the test does not exceed 2⋅105. The sum of the values of m for all sets of input data in the test does not exceed 5⋅105
.
Output
Print t
integers — the answers to the given test cases in the order they are written in the input.
Example
Input
3 7 7 3 5 1 2 2 3 3 4 4 5 5 6 6 7 7 5 4 5 2 3 1 2 2 3 3 4 4 1 4 2 4 3 2 1 1 2 2 3 4 1
Output
4 0 1
Sponsor
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 501050;
ll fa[maxn];
ll Find(ll x)
{
if(fa[x]==x) return x;
else return fa[x]=Find(fa[x]);
}
void Merge(ll x,ll y)
{
x=Find(x);y=Find(y);
if(x!=y) fa[x]=y;
}
map<ll,ll>mp;
vector<ll>a1;vector<ll>b1;
ll visa[maxn],visb[maxn];
int main()
{
ll t;
scanf("%lld",&t);
while(t--)
{
ll n,m,a,b;
scanf("%lld %lld %lld %lld",&n,&m,&a,&b);
a1.clear();b1.clear();
for(int i=1; i<=n+10; i++) fa[i]=i,visa[i]=0,visb[i]=0;
for(ll i=1; i<=m; i++)
{
ll u,v;
scanf("%lld %lld",&u,&v);
if((u==a&&v==b)||(u==b&&v==a)) continue;
if(u==a)a1.push_back(v);
if(v==a)a1.push_back(u);
if(u==b)b1.push_back(v);
if(v==b)b1.push_back(u);
if(u!=a&&u!=b&&v!=a&&v!=b) Merge(u,v);
}
for(auto i:a1)
{
visa[Find(i)]=1;
//printf("i=%lld visa[%lld]=%lld\n",i,Find(i),visa[Find(i)]);
}
for(auto i:b1)
{
visb[Find(i)]=1;
//printf("i=%lld visb[%lld]=%lld\n",i,Find(i),visb[Find(i)]);
}
for(ll i=1;i<=n;i++)
{
visa[i]=visa[Find(i)];
visb[i]=visb[Find(i)];
//printf("visa[%lld]=%lld visb[%lld]=%lld\n",i,visa[i],i,visb[i]);
}
ll x=0,y=0,ans=0;
for(ll i=1;i<=n;i++)
{
if(visa[i]==1&&visb[i]==0) x++;
if(visa[i]==0&&visb[i]==1) y++;
}
ans=x*y;
printf("%lld\n",ans);
}
}