题解 铺地毯
尝试扫描线，没调出来
考完才发现我试图用线段树的pushup完成对值域的离散化
这样极为难写
扫描线上的所有点会将线分成几个形如 \([y_i, y_{i+1}]\) 的区间
令一个 \(c_i\) 为区间 \([y_i, y_{i+1}]\) 被覆盖的次数，维护这个东西即可
但是还有一个 \(O(n)\) 做法
尝试求出 \(n-1\) 个矩形的面积交的并
发现 \(n-1\) 个矩形的面积交可以用一个前缀交和一个后缀交实现
于是 \(n-1\) 个矩形的面积交的重叠部分就是所有矩形的交，减去即可
复杂度 \(O(n)\)
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 600010
#define ll long long
#define fir first
#define sec second
#define make make_pair
#define pb push_back
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, p, q;
int a[N], b[N], c[N], d[N];
namespace force{
int cnt[100][100];
void solve() {
int ans=0;
memset(cnt, 0, sizeof(cnt));
for (int i=1; i<=n; ++i) {
int a=read(), b=read(), c=read()-1, d=read()-1;
for (int j=a; j<=c; ++j) for (int k=b; k<=d; ++k) ++cnt[j][k];
}
for (int i=0; i<100; ++i) for (int j=0; j<100; ++j) if (cnt[i][j]>=n-1) ++ans;
printf("%d\n", ans);
}
}
namespace task1{
#undef unix
int unix[N], usizx, uniy[N], usizy;
vector<pair<int, int>> add[N], del[N];
int tl[N<<2], tr[N<<2], fir[N<<2], fcnt[N<<2], sec[N<<2], scnt[N<<2], tag[N<<2];
#define tl(p) tl[p]
#define tr(p) tr[p]
#define dat(p) dat[p]
void pushup(int p) {
int max1=0, max2=0;
max1=max(fir[p<<1], fir[p<<1|1]);
if (fir[p<<1]!=max1) max2=max(max2, fir[p<<1]);
if (fir[p<<1|1]!=max1) max2=max(max2, fir[p<<1|1]);
max2=max(max2, max(sec[p<<1], sec[p<<1|1]));
fir[p]=max1; sec[p]=max2; fcnt[p]=scnt[p]=0;
if (fir[p<<1]==max1) fcnt[p]+=fcnt[p<<1];
if (fir[p<<1|1]==max1) fcnt[p]+=fcnt[p<<1|1];
if (fir[p<<1]==max2) scnt[p]+=fcnt[p<<1];
if (fir[p<<1|1]==max2) scnt[p]+=fcnt[p<<1|1];
if (sec[p<<1]==max2) scnt[p]+=scnt[p<<1];
if (sec[p<<1|1]==max2) scnt[p]+=scnt[p<<1|1];
}
void spread(int p) {
if (!tag[p]) return ;
fir[p<<1]+=tag[p]; tag[p<<1]+=tag[p]; if (sec[p<<1]) sec[p<<1]+=tag[p];
fir[p<<1|1]+=tag[p]; tag[p<<1|1]+=tag[p]; if (sec[p<<1|1]) sec[p<<1|1]+=tag[p];
tag[p]=0;
}
void build(int p, int l, int r) {
tl(p)=l; tr(p)=r; fir[p]=fcnt[p]=sec[p]=scnt[p]=tag[p]=0;
if (l==r) {fcnt[p]=scnt[p]=1; return ;}
int mid=(l+r)>>1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
pushup(p);
}
void upd(int p, int l, int r, int val) {
if (l<=tl(p) && r>=tr(p)) {
cout<<"at: "<<p<<' '<<tl(p)<<' '<<tr(p)<<' '<<fir[p]<<' '<<fcnt[p]<<endl;
fir[p]+=val; tag[p]+=val;
if (sec[p]) sec[p]+=val;
return ;
}
spread(p);
int mid=(tl(p)+tr(p))>>1;
if (l<=mid) upd(p<<1, l, r, val);
if (r>mid) upd(p<<1|1, l, r, val);
pushup(p);
}
void solve() {
ll ans=0;
for (int i=1; i<=n; ++i) {
a[i]=read(); b[i]=read(); c[i]=read()-1; d[i]=read()-1;
unix[++usizx]=a[i]; unix[++usizx]=c[i];
uniy[++usizy]=b[i]; uniy[++usizy]=d[i];
}
// sort(unix+1, unix+usizx+1); sort(uniy+1, uniy+usizy+1);
// usizx=unique(unix+1, unix+usizx+1)-unix-1; usizy=unique(uniy+1, uniy+usizy+1)-uniy-1;
build(1, 1, q);
for (int i=1; i<=max(p, q); ++i) add[i].clear(), del[i].clear();
for (int i=1; i<=n; ++i) {
// add[lower_bound(unix+1, unix+usizx+1, a[i])-unix].pb(make(lower_bound(uniy+1, uniy+usizy+1, b[i])-uniy, lower_bound(uniy+1, uniy+usizy+1, d[i])-uniy));
// cout<<"add: "<<lower_bound(unix+1, unix+usizx+1, a[i])-unix<<endl;
// del[lower_bound(unix+1, unix+usizx+1, c[i])-unix].pb(make(lower_bound(uniy+1, uniy+usizy+1, b[i])-uniy, lower_bound(uniy+1, uniy+usizy+1, d[i])-uniy));
add[a[i]].pb(make(b[i], d[i]));
del[c[i]].pb(make(b[i], d[i]));
}
for (int i=1; i<=10; ++i) {
cout<<"i: "<<i<<endl;
for (auto it:add[i]) upd(1, it.fir, it.sec, 1), cout<<"upd: "<<it.fir<<' '<<it.sec<<endl;
cout<<"fir: "<<fir[1]<<' '<<fcnt[1]<<endl;
cout<<"sec: "<<sec[1]<<' '<<scnt[1]<<endl;
if (fir[1] && fir[1]>=n-1) ans+=fcnt[1];
if (sec[1] && sec[1]>=n-1) ans+=scnt[1];
for (auto it:del[i]) upd(1, it.fir, it.sec, -1);
}
printf("%lld\n", ans);
}
}
namespace task2{
#undef unix
int unix[N], usizx, uniy[N], usizy;
vector<pair<int, int>> add[N], del[N];
int tl[N<<2], tr[N<<2], fir[N<<2], fcnt[N<<2], sec[N<<2], scnt[N<<2], tag[N<<2];
#define tl(p) tl[p]
#define tr(p) tr[p]
#define dat(p) dat[p]
void pushup(int p) {
int max1=0, max2=0;
max1=max(fir[p<<1], fir[p<<1|1]);
if (fir[p<<1]!=max1) max2=max(max2, fir[p<<1]);
if (fir[p<<1|1]!=max1) max2=max(max2, fir[p<<1|1]);
max2=max(max2, max(sec[p<<1], sec[p<<1|1]));
fir[p]=max1; sec[p]=max2; fcnt[p]=scnt[p]=0;
if (fir[p<<1]==max1) fcnt[p]+=fcnt[p<<1];
if (fir[p<<1|1]==max1) fcnt[p]+=fcnt[p<<1|1];
if (fir[p<<1]==max2) scnt[p]+=fcnt[p<<1];
if (fir[p<<1|1]==max2) scnt[p]+=fcnt[p<<1|1];
if (sec[p<<1]==max2) scnt[p]+=scnt[p<<1];
if (sec[p<<1|1]==max2) scnt[p]+=scnt[p<<1|1];
}
void spread(int p) {
if (!tag[p]) return ;
fir[p<<1]+=tag[p]; tag[p<<1]+=tag[p]; if (sec[p<<1]) sec[p<<1]+=tag[p];
fir[p<<1|1]+=tag[p]; tag[p<<1|1]+=tag[p]; if (sec[p<<1|1]) sec[p<<1|1]+=tag[p];
tag[p]=0;
}
void build(int p, int l, int r) {
tl(p)=l; tr(p)=r; fir[p]=fcnt[p]=sec[p]=scnt[p]=tag[p]=0;
int mid=(l+r)>>1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
pushup(p);
}
void upd(int p, int l, int r, int val) {
if (l<=tl(p) && r>=tr(p)) {
cout<<"at: "<<p<<' '<<tl(p)<<' '<<tr(p)<<' '<<fir[p]<<' '<<fcnt[p]<<endl;
fir[p]+=val; tag[p]+=val;
if (sec[p]) sec[p]+=val;
return ;
}
spread(p);
int mid=(tl(p)+tr(p))>>1;
if (l<=mid) upd(p<<1, l, r, val);
if (r>mid) upd(p<<1|1, l, r, val);
pushup(p);
}
void solve() {
ll ans=0;
for (int i=1; i<=n; ++i) {
a[i]=read(); b[i]=read(); c[i]=read()-1; d[i]=read()-1;
unix[++usizx]=a[i]; unix[++usizx]=c[i];
uniy[++usizy]=b[i]; uniy[++usizy]=d[i];
}
// sort(unix+1, unix+usizx+1); sort(uniy+1, uniy+usizy+1);
// usizx=unique(unix+1, unix+usizx+1)-unix-1; usizy=unique(uniy+1, uniy+usizy+1)-uniy-1;
build(1, 1, usizy);
for (int i=1; i<=usizx; ++i) add[i].clear(), del[i].clear();
for (int i=1; i<=n; ++i) {
// add[lower_bound(unix+1, unix+usizx+1, a[i])-unix].pb(make(lower_bound(uniy+1, uniy+usizy+1, b[i])-uniy, lower_bound(uniy+1, uniy+usizy+1, d[i])-uniy));
// cout<<"add: "<<lower_bound(unix+1, unix+usizx+1, a[i])-unix<<endl;
// del[lower_bound(unix+1, unix+usizx+1, c[i])-unix].pb(make(lower_bound(uniy+1, uniy+usizy+1, b[i])-uniy, lower_bound(uniy+1, uniy+usizy+1, d[i])-uniy));
}
for (int i=1; i<=usizx; ++i) {
cout<<"i: "<<i<<endl;
for (auto it:add[i]) upd(1, it.fir, it.sec, 1), cout<<"upd: "<<it.fir<<' '<<it.sec<<endl;
cout<<"fir: "<<fir[1]<<' '<<fcnt[1]<<endl;
cout<<"sec: "<<sec[1]<<' '<<scnt[1]<<endl;
if (fir[1]>=n-1) ans+=fcnt[1];
if (sec[1]>=n-1) ans+=scnt[1];
for (auto it:del[i]) upd(1, it.fir, it.sec, -1);
}
printf("%lld\n", ans);
}
}
namespace task{
ll ans;
struct rec{
int a, b, c, d;
rec(){}
rec(int w, int x, int y, int z) {a=w; b=x; c=y; d=z;}
inline void build() {a=read(); b=read(); c=read()-1; d=read()-1;}
inline void build(int w, int x, int y, int z) {a=w; b=x; c=y; d=z;}
inline ll s() {
if (a>c || b>d) return 0;
else return 1ll*(c-a+1)*(d-b+1);
}
}r[N], cap, sum[N], suf[N];
inline rec operator & (rec a, rec b) {return rec(max(a.a, b.a), max(a.b, b.b), min(a.c, b.c), min(a.d, b.d));}
void solve() {
for (int i=1; i<=n; ++i) r[i].build();
if (n==1) {printf("%lld\n", r[1].s()); return ;}
cap=r[1]; ans=0;
for (int i=2; i<=n; ++i) cap=cap&r[i];
sum[1]=r[1]; suf[n]=r[n];
for (int i=2; i<=n; ++i) sum[i]=sum[i-1]&r[i];
for (int i=n-1; i; --i) suf[i]=suf[i+1]&r[i];
// cout<<"sum: "; for (int i=1; i<=n; ++i) cout<<sum[i].s()<<' '; cout<<endl;
// cout<<"suf: "; for (int i=1; i<=n; ++i) cout<<suf[i].s()<<' '; cout<<endl;
ans+=suf[2].s()-cap.s();
ans+=sum[n-1].s()-cap.s();
for (int i=2; i<n; ++i) ans+=(sum[i-1]&suf[i+1]).s()-cap.s(); //, cout<<"add: "<<(sum[i-1]&suf[i+1]).s()<<endl;
ans+=cap.s();
printf("%lld\n", ans);
}
}
signed main()
{
freopen("carpet.in", "r", stdin);
freopen("carpet.out", "w", stdout);
int T=read();
while (T--) {
p=read(); q=read(); n=read();
// force::solve();
// task1::solve();
task::solve();
}
return 0;
}
