如果不考虑坐标范围可以显然的得到一个的
实际上可以发现对于一个位置,
如果一个洞不是和他的曼哈顿距离小于等于2,那么就一定不会被这个洞更新
并且除此之外就只可能会被左下方的点更新
因此只需要把所有特殊更新的点提出来
按对角线用维护转移即可
#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ib==ob)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0;bool f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b,a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(int x){return (x<0)?(x+mod):x;}
cs int N=100005;
map<pii,int>val;
map<int,int>f;
map<int,int>pos;
int n,m,k,tot,ans,buc[N<<3];
pii p[N<<3];
#define It map<int,int>::iterator
struct comp{
inline bool operator ()(cs pii &a,cs pii &b){
return (a.fi+a.se==b.fi+b.se)?(a.fi<b.fi):(a.fi+a.se>b.fi+b.se);
}
};
inline int calc(int x,int y){
if(val.count(pii(x,y)))return val[pii(x,y)];
int v1=0,v2=0;
if(val.count(pii(x+1,y)))v1=val[pii(x+1,y)];
else v1=max(f[x-y],f[x-y+2]);
if(val.count(pii(x,y+1)))v2=val[pii(x,y+1)];
else v2=max(f[x-y],f[x-y-2]);
return min(v1,v2);
}
int main(){
#ifdef Stargazer
freopen("lx.in","r",stdin);
#endif
n=read(),m=read(),k=read();
for(int i=1;i<=k;i++){
int x=read(),y=read(),v=read();
val[pii(x,y)]=v;
for(int k=0;k<=2&&x-k>0;k++)
for(int j=0;k+j<=2&&y-j>0;j++)
p[++tot]=pii(x-k,y-j);
}
sort(p+1,p+tot+1,comp());
tot=unique(p+1,p+tot+1)-p-1;
for(int i=1,nxt;i<=tot;i=nxt+1){
for(nxt=i;nxt<tot&&p[nxt].fi+p[nxt].se==p[nxt+1].fi+p[nxt+1].se;nxt++);
for(int j=i;j<=nxt;j++){
int x=p[j].fi,y=p[j].se;
if(f.count(x-y))Add(ans,fix(1ll*f[x-y]*(p[pos[x-y]].fi-x)%mod));
buc[j]=calc(x,y);
}
for(int j=i;j<=nxt;j++){
int x=p[j].fi,y=p[j].se;
f[x-y]=buc[j],pos[x-y]=j;
}
}
for(It it=f.bg();it!=f.end();it++){
Add(ans,fix(1ll*it->se*min(p[pos[it->fi]].fi,p[pos[it->fi]].se)%mod));
}
cout<<ans<<'
';
}