C++(clang++11) 解法, 执行用时: 327ms, 内存消耗: 31776K, 提交时间: 2021-02-14 15:50:51

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include <cstdlib>
using namespace std;
#define maxk 300005
long long read()
{
    long long x=0,f=1; char ch=getchar();
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
 
long long max(long long a,int b)
{
    if (a>b) return a;
    return b;
}
struct data{long long x,y,r;}in[maxk];
struct dat{
    long long x,y1,y2,k;
    bool operator < (const dat & A) const
        {
            return x<A.x;
        }
};
struct da{
    long long st,ed;
    long long sqr()
        {
            if ((ed-st+1)&1) return (ed-st+2)*((ed-st+1)/2+1)/2;
            else return (ed-st+1+2)*((ed-st+1-2)/2+1)/2;
        }
    bool operator < (const da A) const
        {
            return st<A.st;
        }
};
int n,m,k;
#define maxn 1200005
struct SegmentTree{
    int del[maxn],tree[maxn];
    int Y[maxk];
    void clear()
        {
            memset(del,0,sizeof(del));
            memset(tree,0,sizeof(tree));
        }
    void updata(int now,long long l,long long r)
        {
            if (del[now]>0) tree[now]=r-l;
            else tree[now]=tree[now<<1]+tree[now<<1|1];
        }
    void insert(int now,int l,int r,int L,int R,int num)
        {
            int mid=(l+r)>>1;
            if (L<=Y[l] && R>=Y[r]) del[now]+=num; else
                {
                    if (L<Y[mid]) insert(now<<1,l,mid,L,R,num);
                    if (R>Y[mid]) insert(now<<1|1,mid,r,L,R,num);
                }
            updata(now,Y[l],Y[r]);
        }
    long long query()
        {
            return tree[1];
        }
}T;
 
struct cal{
    dat line[maxk];
    int tot,n;
    long long x1[maxk],x2[maxk],y1[maxk],y2[maxk];
    long long calc()
        {
            T.clear();
            for (int i=1; i<=n; i++) T.Y[i*2-1]=y1[i],T.Y[i*2]=y2[i];
            sort(T.Y+1,T.Y+2*n+1); tot=1;
            for (int i=2; i<=2*n; i++) if (T.Y[i]!=T.Y[i-1]) T.Y[++tot]=T.Y[i];
            for (int i=1; i<=n; i++)
                {
                    line[i*2-1].x=x1[i]; line[i*2-1].y1=y1[i]; line[i*2-1].y2=y2[i];
                    line[i*2].x=x2[i]; line[i*2].y1=y1[i]; line[i*2].y2=y2[i];
                    line[i*2-1].k=1; line[i*2].k=-1;
                }
            sort(line+1,line+2*n+1);
            long long ans=0,last=0;
            for (int i=1; i<=2*n; i++)
                {
                    if (i!=1) ans=ans+last*(long long)((long long)(line[i].x)-(long long)line[i - 1].x);
                    if (last<0) {int a;a+=1;}
                    if (line[i].y1!=line[i].y2) T.insert(1,1,tot,line[i].y1,line[i].y2,line[i].k);
                    last=T.query();
                }
        return ans;
    }
}calc1, calc2;
struct calcc{
    int n;
    da tri[maxk],tmp;
    long long cal()
        {
            if (n==0) return 0ll;
            long long ans=0; da tmp;
            sort(tri+1,tri+n+1);
            ans=tri[1].sqr(); int now=1;
            for (int i=2; i<=n; i++)
                {
                    if (tri[i].st>=tri[now].st && tri[i].ed<=tri[now].ed) continue;
                    if (tri[i].st>tri[now].ed)
                        {
                            ans+=tri[i].sqr(); now=i; continue;
                        }
                    ans+=tri[i].sqr();tmp.st=tri[i].st; tmp.ed=tri[now].ed;
                    ans-=tmp.sqr(); now=i;
                }
            return ans;
        }
}tc[5];
 
int main()
{
    n=read(),m=read(),k=read();
    for (int i=1; i<=k; i++) in[i].x=read(),in[i].y=read(),in[i].r=read();
    calc1.n=calc2.n=k;
    for (int i=1; i<=k; i++)
        {
            int tmp=in[i].r-((in[i].x+in[i].y+in[i].r)&1);
            calc1.x1[i]=(in[i].x+in[i].y-tmp)>>1;
            calc1.y1[i]=(in[i].y-tmp-in[i].x)>>1;
            calc1.x2[i]=((in[i].x+in[i].y+ tmp)>>1)+1;
            calc1.y2[i]=((in[i].y+tmp-in[i].x)>>1)+1;
        }
    for (int i=1; i<=k; i++)
        {
            int tmp=in[i].r-(!((in[i].x+in[i].y+in[i].r)&1));
            calc2.x1[i]=(in[i].x+in[i].y-tmp-1)>>1;
            calc2.y1[i]=(in[i].y-tmp-in[i].x-1)>>1;
            calc2.x2[i]=((in[i].x+in[i].y+tmp-1)>>1)+1;
            calc2.y2[i]=((in[i].y+tmp-in[i].x-1)>>1)+1;
        }
    long long ans=0;
    ans=calc1.calc()+calc2.calc();
    for (int i=1; i<=k; i++) if (in[i].r>=in[i].x)
        {
            ++tc[1].n;
            tc[1].tri[tc[1].n].st=in[i].y-(in[i].r-in[i].x);
            tc[1].tri[tc[1].n].ed=in[i].y+(in[i].r-in[i].x);
        }
    for (int i=1; i<=k; i++) if (in[i].r>=in[i].y)
        {
            ++tc[2].n;
            tc[2].tri[tc[2].n].st=in[i].x-(in[i].r-in[i].y);
            tc[2].tri[tc[2].n].ed=in[i].x+(in[i].r-in[i].y);
        }
    for (int i=1; i<=k; i++) if (in[i].r>=n+1-in[i].x)
        {
            ++tc[3].n;
            tc[3].tri[tc[3].n].st=in[i].y-(in[i].r-(n+1-in[i].x));
            tc[3].tri[tc[3].n].ed=in[i].y+(in[i].r-(n+1-in[i].x));
        }
    for (int i=1; i<=k; i++) if (in[i].r>=m+1-in[i].y)
        {
            ++tc[4].n;
            tc[4].tri[tc[4].n].st=in[i].x-(in[i].r-(m+1-in[i].y));
            tc[4].tri[tc[4].n].ed=in[i].x+(in[i].r-(m+1-in[i].y));
        }
    for (int i=1; i<=4; i++)
        ans-=tc[i].cal();
    long long m1=0,m2=0,m3=0,m4=0;
    for (int i=1; i<=k; i++)
        {
            if (in[i].r>=in[i].x+in[i].y) m1=max(m1,1+in[i].r-in[i].x-in[i].y);
            if (in[i].r>=in[i].x+(m+1-in[i].y)) m2=max(m2, 1+in[i].r-in[i].x-(m+1-in[i].y));
            if (in[i].r>=in[i].y+(n+1-in[i].x)) m3=max(m3, 1+in[i].r-in[i].y-(n+1-in[i].x));
            if (in[i].r>=(m+1-in[i].y)+(n+1-in[i].x)) m4=max(m4,1+in[i].r-((m+1-in[i].y)+(n+1-in[i].x)));
        }
    ans+=m1*(m1+1)/2+m2*(m2+1)/2+m3*(m3+1)/2+m4*(m4+1)/2;
    printf("%lld\n",ans);
    return 0;
}