the problem is at first complicated, but once you hold the concept of 2d array prefix and 2d difference array you will find it simple, first you have to swap coordinates such that x1<x2 and y1<y2 and after that you construct the difference array such that each rectangle is filled with a bunch of nonzero numbers , then you construct the 2d prefix array, finally we count all zero's in the prefix where 0 means not surrounded by a rectangle and any non zero number indicates the opposite

include <bits/stdc++.h>

include

using namespace std;

define fre freopen("input.txt","r",stdin); freopen("output.txt","w",stdout)

define read(t) ll t; cin>>t

define reads(s) string s; cin>>s

define Fast ios_base::sync_with_stdio(false);cin.tie(NULL)

define once solve()

define much(t) ll t; cin>>t; while(t--)solve()

define el '\n'

define ll long long

define vl(v,n) vectorv(n)

define yes cout<<"YES"<<el

define no cout<<"NO"<<el

define ins(vec,n) vectorvec(n);for(auto&i:vec)cin>>i

define all(s) s.begin(),s.end()

define up(i,a,n) for(ll i=a;i<n;i++)

define down(i,a,n) for(ll i=a;i>=n;i--)

define put(v) for(auto i:v) cout<<i<<' '; cout<<el

define put2d(v) up(i,0,v.size()){ up(j,0,v[i].size()){cout<<v[i][j]<<' ';} cout<<el;}

define putpair(v) for(auto i:v)cout<<"("<<i.first<<","<<i.second<<") "; cout<<el

const int Max = 200007; const int MOD = 1000000007; ll tc = 1; void solve() { read(w); read(h); read(n); vector<vector>diffArr(w+1,vector(h+1,0)); vector<vector>prefix(w+1,vector(h+1,0)); while(n--) { read(x1);read(y1);read(x2);read(y2); if (x1 > x2)swap(x1, x2); if (y1 > y2)swap(y1, y2); diffArr[x1-1][y1-1] += 1; diffArr[x2][y1-1] -= 1; diffArr[x1-1][y2] -= 1; diffArr[x2][y2] += 1; } up(i,1,w+1){ up(j,1,h+1){ prefix[i][j]=diffArr[i-1][j-1]+prefix[i-1][j]+prefix[i][j-1]-prefix[i-1][j-1]; } } ll cnt{}; up(i,1,w+1){ up(j,1,h+1){ if(prefix[i][j]==0)cnt++; } } cout<<cnt<<el; }

int main() { Fast;

ifndef ONLINE_JUDGE

fre;

endif

if(tc){much(t);}
else {once;}
return 0;

}