Vasya and Magic Matrix CodeForces - 1042E (概率dp)

Posted 2021-12-02 uid001

大意:给定n*m矩阵, 初始位置(r,c), 每一步随机移动到权值小于当前点的位置, 得分为移动距离的平方, 求得分期望.

 

直接暴力dp的话复杂度是O(n^4), 把距离平方拆开化简一下, 可以O(n^2logn).

 

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P2%x)*(P2-P2/x)%P2;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



const int N = 1e3+10;
int n, m, r, c, cnt;
struct _ {
	int x,y,w;
	bool operator < (const _ &rhs) const {
		return w<rhs.w;
	}
} a[N*N];
int dp[N][N];

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) REP(j,1,m) { 
		int t;
		scanf("%d", &t);
		a[++cnt] = {i,j,t};
	}
	sort(a+1,a+1+cnt);
	a[cnt+1].w=-1;
	ll sum_dp = 0, sum_2 = 0, sum_x = 0, sum_y = 0;
	REP(R,1,cnt) {
		int L = R;
		while (a[R].w==a[R+1].w) ++R;
		if (L!=1) {
			REP(i,L,R) {
				ll t = (sum_dp+sum_2-a[i].x*sum_x-a[i].y*sum_y+(L-1)*((ll)a[i].x*a[i].x+(ll)a[i].y*a[i].y))%P2;
				t = t*inv(L-1)%P2;
				if (t<0) t += P2;
				dp[a[i].x][a[i].y] = t%P2;
			}
		}
		REP(i,L,R) {
			(sum_dp += dp[a[i].x][a[i].y]) %= P2;
			(sum_2 += (ll)a[i].x*a[i].x+(ll)a[i].y*a[i].y) %= P2;
			(sum_x += 2*a[i].x) %= P2;
			(sum_y += 2*a[i].y) %= P2;
		}
	}
	scanf("%d%d", &r, &c);
	printf("%d\n", dp[r][c]);
}