NOIP 模拟 6 辣鸡

\(AC\kern 0.4em CODE:\)

#include<bits/stdc++.h>
#define ri register int
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
    inline int read() {
        ri x=0,f=1;char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        return x*f; 
    }
}
using IO::read;
namespace nanfeng{
    #define int long long
    #define cmax(x,y) ((x)>(y)?(x):(y))
    #define cmin(x,y) ((x)>(y)?(y):(x))
    #define FI FILE *IN
    #define FO FILE *OUT
    #undef bool
    static const int N=1e5+7;
    struct Matrix{
        int x1,y1,x2,y2;
        friend inline bool operator<(Matrix m1,Matrix m2) {return m1.x1<m2.x1;}
        Matrix(){}
        Matrix(int x1,int y1,int x2,int y2):x1(x1),y1(y1),x2(x2),y2(y2){}
    }mat[N];
    int n,ans;
    inline int calc(Matrix m) {
        int res=0;
        int x=m.x2-m.x1,y=m.y2-m.y1;
        if (x>y) swap(x,y);
        res+=(x*x<<1);
        res+=(y-x)*(x<<1);
        return res;
    }
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        n=read();
        for (ri i(1);i<=n;p(i)) {
            int x1=read(),y1=read(),x2=read(),y2=read();
            mat[i]=Matrix(x1,y1,x2,y2);
        }
        if (n==1) {printf("%lld\n",calc(mat[1]));return 0;}
        sort(mat+1,mat+n+1);
        // for (ri i(1);i<=n;p(i)) 
        for (ri i(1);i<n;p(i)) {
            ans+=calc(mat[i]);
            // printf("%lld %lld %lld %lld\n",mat[i].x1,mat[i].y1,mat[i].x2,mat[i].y2);
            for (ri j(i+1);j<=n;p(j)) {
                if (mat[j].x1-mat[i].x2>1) break;
                if (mat[j].x1-mat[i].x2==1) {
                    if (mat[j].y1-mat[i].y2>1||mat[i].y1-mat[j].y2>1) continue;
                    if (mat[j].y1-mat[i].y2==1||mat[i].y1-mat[j].y2==1) {ans+=1;continue;}
                    ans+=(cmin(mat[i].y2,mat[j].y2)-cmax(mat[i].y1,mat[j].y1))<<1;
                    if (cmax(mat[i].y2,mat[j].y2)>cmin(mat[i].y2,mat[j].y2)) ans+=1;
                    if (cmax(mat[i].y1,mat[j].y1)>cmin(mat[i].y1,mat[j].y1)) ans+=1;
                } else if (mat[j].y1>mat[i].y2) {
                    if (mat[j].y1-mat[i].y2>1) continue;
                    ans+=(cmin(mat[i].x2,mat[j].x2)-mat[j].x1)<<1;
                    if (mat[i].x1<mat[j].x1) ans+=1;
                    if (cmax(mat[i].x2,mat[j].x2)>cmin(mat[i].x2,mat[j].x2)) ans+=1;
                } else {
                    if (mat[i].y1-mat[j].y2>1) continue;
                    ans+=(cmin(mat[i].x2,mat[j].x2)-mat[j].x1)<<1;
                    if (mat[i].x1<mat[j].x1) ans+=1;
                    if (cmax(mat[i].x2,mat[j].x2)>cmin(mat[i].x2,mat[j].x2)) ans+=1;
                }
                // if (i==2) printf("%lld %lld %lld %lld ans=%lld\n",mat[j].x1,mat[j].y1,mat[j].x2,mat[j].y2,ans);
            } 
            // printf("ans=%lld\n",ans);
        }
        // printf("%lld %lld %lld %lld\n",mat[n].x1,mat[n].y1,mat[n].x2,mat[n].y2);
        printf("%lld\n",ans+calc(mat[n]));
        return 0;
    }
    #undef int
}
int main() {return nanfeng::main();}