題目連結 一道期望DP題，先把矩陣的元素從小到大排序。f[i]表示期望 $f[i]=\frac{\sum_{j=1}^{t}{(f[j]+(x[i]-x[j])^2+(y[i]-y[j])^2})}{t}$ 然後把式子轉換一下 $f[i]=\frac{\sum{(f[j]+x[j]^2+y[j]^2)}-2 x[i]\times \sum x[j]-2y[i]\times \sum y[j]}{t}+x[i]^2+y[i]^2$

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int mo=998244353;
const int N=1001;
struct node{int x,y;}p[N*N];
int
 
 X,Y,n,m,a[N][N],tot,V;
long long f,fs,xs,ys,fsw,xsw,ysw;//f[i]=(sigma(f[j]+(x[i]-x[j])^2+(y[i]-y[j])^2))/p
int getint();//f[i]=(sigma(f[j]+x[j]^2+y[j]^2)-2*x[i]*sigma(x[j])-2*y[i]*sigma(y[j]))/p+x[i]^2+y[i]^2
bool cmp(node x,node y){return a[x.x][x.y]<a[y.x][y.y];}
long long ks(long long t,long long k)
 
{
	long long w;
	for(w=1;k;k>>=1,t=t*t%mo)if(k&1)w=w*t%mo;
	return w;
}
int main(){
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	int i,j,k,t;
	n=getint();m=getint();
	for(i=1;i<=n;i++)for(j=1;j<=m;j++){
	    a[i][j]=getint();
	    p[++tot].x=i;p[tot].y=j;
	}
	X=getint();Y=getint();
	sort(p+1,p+tot+1,cmp);
	for(i=1;i<=tot;i++)if(p[i].x==X&&p[i].y==Y){V=i;break;}
	for(i=1,j=0;i<=V;i++){
		f=fs-2*p[i].x*xs%mo+-2*p[i].y*ys%mo+mo+mo;
		f%=mo;
		f=f*ks(j,mo-2)%mo;
		f+=p[i].x*p[i].x%mo*ks(j,mo-1);f%=mo;
		f+=p[i].y*p[i].y%mo*ks(j,mo-1);f%=mo;
		fsw+=f;fsw+=p[i].x*p[i].x%mo;fsw+=p[i].y*p[i].y%mo;fsw%=mo;
		xsw+=p[i].x;xs%=mo;
		ysw+=p[i].y;ys%=mo;
		if(a[p[i].x][p[i].y]<a[p[i+1].x][p[i+1].y]){
			j=i;
			fs+=fsw;fs%=mo;
			xs+=xsw;xs%=mo;
			ys+=ysw;ys%=mo;
			fsw=xsw=ysw=0;
		}
	}
	printf("%lld",f);
}
int getint(){
	int w=0,q=1;
	char c=getchar();
	while((c>'9'||c<'0')&&c!='-')c=getchar();
	if(c=='-')q=-1,c=getchar();
	while(c>='0'&&c<='9')w=w*10+c-'0',c=getchar();
	return w*q;
}