链接:http://codeforces.com/gym/101982/attachments
思路:
问被覆盖次数为奇数次的矩阵的面积并
扫描线求矩阵面积并我们是上界赋为-1,下界赋为1,因为要求覆盖次数为奇数次的,我们直接上下界都赋值为1,然后每次区间更新的时候对这段区间取异或就好了
实现代码;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid ll m = (l + r) >> 1
const ll M = 2e5+;
struct seg{
ll l,r,h;
ll s;
seg(){}
seg(ll a,ll b,ll c,ll d):l(a),r(b),h(c),s(d){}
bool operator < (const seg &cmp) const {
return h < cmp.h;
}
}t[M];
ll sum[M<<],x[M<<];
ll cnt[M<<];
void pushup(ll rt){
sum[rt] = sum[rt<<] + sum[rt<<|];
} void pushdown(ll l,ll r,ll rt){
if(cnt[rt]){
mid;
sum[rt<<] = x[m] - x[l-] - sum[rt<<];
sum[rt<<|] = x[r] - x[m] - sum[rt<<|];
cnt[rt<<] ^= ;
cnt[rt<<|] ^= ;
cnt[rt] = ;
}
} void update(ll L,ll R,ll c,ll l,ll r,ll rt){
if(L <= l&&R >= r){
cnt[rt] ^= ;
sum[rt] = (x[r]-x[l-]) - sum[rt];
return ;
}
pushdown(l,r,rt);
mid;
if(L <= m) update(L,R,c,lson);
if(R > m) update(L,R,c,rson);
pushup(rt);
} ll bin(ll key,ll n,ll x[]){
ll l = ;ll r = n-;
while(l <= r){
mid;
if(x[m] == key) return m;
else if(x[m] < key) l = m+;
else r = m-;
}
return -;
}
int main()
{
ll n,cas = ;
ll a,b,c,d;
cin>>n;
ll m = ;
while(n--){
cin>>a>>b>>c>>d;
x[m] = a;
t[m++] = seg(a,c,b,);
x[m] = c;
t[m++] = seg(a,c,d,);
}
sort(x,x+m);
sort(t,t+m);
ll nn = ;
for(ll i = ;i < m;i++){
if(x[i]!=x[i-]) x[nn++] = x[i];
}
ll ret = ;
for(ll i = ;i < m-;i ++){
ll l = bin(t[i].l,nn,x);
ll r = bin(t[i].r,nn,x);
if(l <= r) update(l+,r,t[i].s,,M,);
ret += sum[] * (t[i+].h - t[i].h);
}
cout<<ret<<endl;
return ;
}