『HGOI 20190917』Lefkaritika 题解 (DP)

# pragma GCC optimize(3)
# include<bits/stdc++.h>
# define int long long
using namespace std;
const int N=2e5+10;
int n,m,q;
namespace fast_IO{
    const int IN_LEN = 10000000, OUT_LEN = 10000000;
    char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
    inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
    inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
    inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
    int read(){
        int x = 0; int zf = 1; char ch = ‘ ‘;
        while (ch != ‘-‘ && (ch < ‘0‘ || ch > ‘9‘)) ch = getchar_();
        if (ch == ‘-‘) zf = -1, ch = getchar_();
        while (ch >= ‘0‘ && ch <= ‘9‘) x = x * 10 + ch - ‘0‘, ch = getchar_(); return x * zf;
    }
    void write(int x){
        if (x < 0) putchar_(‘-‘), x = -x;
        if (x > 9) write(x / 10);
        putchar_(x % 10 + ‘0‘);
    }
}
set< pair<int,int> >st;
struct point { int x,y;}r[N];
int cross(point a,point b) {
    return a.x*b.y-a.y*b.x;
}
int direct (point a,point b,point c) {
    point ba={a.x-b.x,a.y-b.y};
    point bc={c.x-b.x,c.y-b.y};
    int ret=cross(bc,ba);
    if (ret==0) return 0;
    else if (ret<0) return -1;
    else if (ret>0) return 1;
}
pair<int,int>bag[N],p[N];
int work(int b) {
    int cnt=0;
    int Lx=max(-b,0ll),Rx=min(n,m-b);
    int Ly=Lx+b,Ry=Rx+b;
    for (int i=1;i<=q;i++) {
        int rec=direct(r[i],(point){-b,0},(point){0,b}); if (b<0) rec=-rec;
        if (b==0) rec=direct(r[i],(point){-1,-1},(point){0,0});
        if (rec==-1) {
            if (r[i].y-b>=Lx&&r[i].y-b<=Rx&&r[i].y>=Ly&&r[i].y<=Ry) bag[++cnt]=make_pair(r[i].y-b,r[i].x-r[i].y+b-1);
        } else if (rec==0) {
            if (r[i].x>=Lx&&r[i].x<=Rx&&r[i].y>=Ly&&r[i].y<=Ry) bag[++cnt]=make_pair(r[i].x,0);
        } else if (rec==1) {
            if (r[i].x>=Lx&&r[i].x<=Rx&&r[i].x+b>=Ly&&r[i].x+b<=Ry) bag[++cnt]=make_pair(r[i].x,r[i].y-r[i].x-b-1);
        }
    }
    sort(bag+1,bag+1+cnt);
    int tot=0;
    for (int i=1,j;i<=cnt;i=j) {
        j=i; int ret=bag[i].second;
        while (bag[j].first==bag[i].first&&j<=cnt) j++;
        p[++tot]=make_pair(bag[i].first,ret);
    }
    if (tot==0) {
        int tmp=min(n-Lx,m-Ly);
        return (tmp+1)*(tmp)/2;
    }
    int ans=(min(n-(p[tot].first+1),m-(p[tot].first+1+b)))*(min(n-(p[tot].first+1),m-(p[tot].first+1+b))+1)/2;
    p[tot].second=min(p[tot].second,min(n-p[tot].first,m-p[tot].first-b));
    int last=-1;
    for (int i=tot;i>=1;i--) {
        if (last==-1) { ans+=p[i].second; last=i; continue;}
        int num=p[last].first-p[i].first-1;
        int val=p[last].second+1;
        if (st.find(make_pair(p[last].first,p[last].first+b))!=st.end()) val--;
        ans+=num*(val+val+num-1)/2;
        if (st.find(make_pair(p[last].first,p[last].first+b))!=st.end()) p[i].second=min(p[i].second,p[last].second+num);
        else p[i].second=min(p[i].second,p[last].second+num+1);
        ans+=p[i].second; 
        last=i;
    }
    int num=p[1].first-Lx;
    int val=p[1].second+1;
    if (st.find(make_pair(p[1].first,p[1].first+b))!=st.end()) val--;
    ans+=num*(val+val+num-1)/2;
    return ans;
}
using namespace fast_IO;
signed main() {
    n=read(),m=read(),q=read(); n--; m--;
    for (int i=1;i<=q;i++) {
        r[i].x=read();r[i].y=read();
        r[i].x--; r[i].y--;
        st.insert(make_pair(r[i].x,r[i].y));
    }
    int ans=0;
    for (int i=-n;i<=m;i++) ans+=work(i);
    write(ans); flush();
    return 0;
}