CF815D Karen and Cards

CF815D Karen and Cards

固定一维c，然后(a,b)看成坐标，矩形区域求交

1.Segment tree Beats！

2.改成不合法的区域就是求并，c反向枚举，区域只增不减且完全包含之前的

单调栈预处理找到轮廓线，然后两个指针维护顶头位置即可

#include<bits/stdc++.h>
#define reg register int
#define il inline
#define int long long
#define numb (ch^'0')
using namespace std;
typedef long long ll;
il void rd(int &x){
    char ch;x=;bool fl=false;
    while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
    for(x=numb;isdigit(ch=getchar());x=x*+numb);
    (fl==true)&&(x=-x);
}
namespace Miracle{
const int N=+;
int n,A,B,C;
ll ans,sum;
struct node{
    int a,b,c;
}p[N];
bool cmpa(node a,node b){
    return a.a<b.a;
}
bool cmpc(node a,node b){
    return a.c>b.c;
}
int sta[N],top;
int x[N],y[N];
int main(){
    rd(n);rd(A);rd(B);rd(C);
    for(reg i=;i<=n;++i) rd(p[i].a),rd(p[i].b),rd(p[i].c);
    sort(p+,p+n+,cmpa);
    for(reg i=;i<=n;++i){
        while(top&&p[sta[top]].b<p[i].b) --top;
        sta[++top]=i;
    }
    sum=(ll)A*B;
    sta[top+]=;
    for(reg i=;i<=top;++i){
        //cout<<" ii "<<i<<" "<<p[sta[i]].b<<endl;
        sum-=(ll)((ll)p[sta[i]].a-p[sta[i-]].a)*p[sta[i]].b;
        for(reg j=p[sta[i-]].a+;j<=p[sta[i]].a;++j) x[j]=p[sta[i]].b;
        for(reg j=p[sta[i]].b;j>p[sta[i+]].b;--j) y[j]=p[sta[i]].a;
    }
//    for(reg i=1;i<=A;++i){
//        cout<<x[i]<<" ";
//    }cout<<endl;
//    for(reg i=1;i<=B;++i){
//        cout<<y[i]<<" ";
//    }cout<<endl;
//    cout<<" sum "<<sum<<endl;
    sort(p+,p+n+,cmpc);
    for(reg tx=,ty=,j=,i=C;i>=;--i){
    //    cout<<" i "<<i<<" tx "<<tx<<" ty "<<ty<<endl;
        while(j<=n&&p[j].c>=i){
        //    cout<<" p[j] "<<j<<" "<<p[j].c<<endl;
            for(;tx<=p[j].a;++tx) sum-=B-max(ty-,x[tx]);
        //    cout<<" sum1 "<<sum<<endl;
            for(;ty<=p[j].b;++ty) sum-=A-max(tx-,y[ty]);
        //    cout<<" sum2 "<<sum<<endl;
            ++j;
        }
        ans+=sum;
    }
    printf("%lld",ans);
    return ;
}

}
signed main(){
    Miracle::main();
    return ;
}

/*
   Author: *Miracle*
   Date: 2019/3/3 15:45:41
*/

很套路了

排序，然后偏序想到坐标矩形求面积

交和并的转化，容斥。或者大力吉司机线段树